[PHP-DB] union/select statement number of columns
Hello all, I receive an error of the following: The used SELECT statements have a different number of columns. Any help, pointers, tutorials are appreciated. Here are the two tables structure I am trying to pull from: Table 1 mysql describe orders; ++--+--+-+-++ | Field | Type | Null | Key | Default | Extra | ++--+--+-+-++ | id | int(255) | NO | PRI | | auto_increment | | ordernum | int(10) | NO | | || | date | varchar(60) | NO | | || | time | varchar(20) | NO | | || | group | varchar(20) | NO | | || | purpose| varchar(255) | NO | | || | tracking | varchar(120) | NO | | || | contact| varchar(255) | NO | | || | eta| varchar(50) | NO | | || | department | varchar(125) | NO | | || | notes | varchar(255) | NO | | || ++--+--+-+-++ 11 rows in set (0.01 sec) Table 2 mysql describe order_items; +-+---+--+-+-++ | Field | Type | Null | Key | Default | Extra | +-+---+--+-+-++ | id | int(11) | NO | PRI | | auto_increment | | ordernum| int(124) | NO | | || | quantity| int(124) | NO | | || | description | varchar(124) | NO | | || | price | decimal(10,0) | NO | | || | partnum | varchar(255) | NO | | || | vendor | varchar(255) | NO | | || +-+---+--+-+-++ 7 rows in set (0.00 sec) And here is the statement I am using (PHP): $query = ( SELECT * FROM `orders` WHERE ( `ordernum` LIKE $var\ OR `purpose` LIKE \$var\ OR `tracking` LIKE \$var\ OR `contact` LIKE \$var\ OR `date` LIKE \$var\ OR `time` LIKE \$var\ OR `eta` LIKE \$var\ OR `department` LIKE \$var\ OR `notes` LIKE \$var\ ) AND `group` = \$group\ ) UNION ( SELECT * FROM `order_items` WHERE ( `ordernum` LIKE \$var\ OR `price` LIKE \$var\ OR `partnum` LIKE \$var\ OR `vendor` LIKE \$var\ OR `quantity` LIKE \$var\ OR `description` LIKE \$var\ ) ORDER BY `ordernum` ); -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Looking for a small open-source PHP-DB app that utilize OOP.
Well some resources for learning it first as these are usually the best
source of information:
Tutorials:
http://www.phpfreaks.com/tutorials.php
http://www.php-editors.com/articles/simple_php_classes.php
Examples:
http://www.phpclasses.org
I really recomend reading and following these as it will undoubtably
help with learning OOP with PHP/MySQL.
As a simple example here is a class, functions and how to use:
?PHP
// class.inc.php
class TestOOP
{
function ChkPHPFuncs( $function_name )
{
if( empty( $function_name ) ) {
return 1;
} else {
if( !function_exists( $function_name ) ) {
return 1;
} else {
return 0;
}
}
}
}
?
To use within a script:
?PHP
//include our class file so lib is available
include 'class.inc.php';
// check for mcrypt functions compiled with php
$list = array( 'mcrypt_cbc', 'mcrypt_cfb', 'mcrypt_create_iv',
'mcrypt_decrypt', 'mcrypt_ecb', 'mcrypt_encrypt' );
// initialize class file so we may use functions from that class
$x = new TestOOP();
//now use $x as our handle for any function from within class file
for( $i = 0; $i count( $list ); $i++ ) {
echo $x-chkPHPFuncs( $list[$i] );
}
?
Also here are some resources from PHP.net for reference:
http://us3.php.net/zend-engine-2.php
http://www.php.net/oop
HTH
Jas
T K wrote:
Hi,
For a study purpose, I'm looking for a small open source program that works
in PHP-MySQL. In particular, I would like to see some examples with PHP 5
and Object-Oriented Programming (such as class, objects, and so on).
I'm looking for a small program, possibly like blog, wiki, forum and so on.
I've downloaded several programs, but most of them don't use
classes/objects.
I'm using PHP without OOP, but recently set out to learn OOP.
Any suggestions would be appreciated.
Tek
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Re: [PHP-DB] Re: mysql statement [SOLVED]
Thanks, most of that I knew but the grouping in your first example is the one that allowed me to only use records matching any field but still limited by the one db field. So thanks again. I was not aware of the parenthesis for grouping objects like a mathematical equation where anything inside the paren's gets executed first and still conditional on the remainder of the equation. TG wrote: Couple of little pointers. If you're doing the sub-select, then you don't need the group like 'mac' because you've already limited your query in the subselect to a specific groupname. The subselect is probably unnecessary since what you're doing is relatively simple and uses the same table. It probably just adds overhead. But in relation to that, you need to segregate your OR statements from your final AND statement or it won't limit by group properly. WHERE ( `ordernum` LIKE 35132 OR `price` LIKE 35132 OR `partnum` LIKE 35132 OR `vendor` LIKE 35132 OR `purpose` LIKE 35132 OR `tracking` LIKE 35132 OR `contact` LIKE 35132 ) AND `group` LIKE 'mac' Notice the ( and ). Next pointer.. LIKE is used when you're doing a non-exact search, but you haven't used any wildcards to indicate a partial search. So what you're essentially doing is the same as price = 35132.If you're doing a multi-field search and want to use LIKE, you'd do this on each: `price` LIKE %35132% The % is sorta like * in other systems. Any quantity of any characters can match that space. Or you can do`price` LIKE 35132% if you want to search the beginning of the field (note the % at the end, but not the beginning of the search string this time). One last thing... I don't know if it increases speed or not, but since you're using so many fields this MAY speed up the query.Depends on how much data you're searching through and if you're doing %search% or need to find start/end strings. Anyway, here's the test.. what's faster, what you're trying to do: WHERE ( `ordernum` LIKE 35132 OR `price` LIKE 35132 OR `partnum` LIKE 35132 OR `vendor` LIKE 35132 OR `purpose` LIKE 35132 OR `tracking` LIKE 35132 OR `contact` LIKE 35132 ) AND `group` LIKE 'mac' or something like this WHERE CONCAT(`ordernum`, `price`, `partnum`, `vendor`, `purpose`, `tracking`, `contact`) LIKE %35132% AND `group` LIKE '%mac%' All the LIKE comparisons against a ton of data MAY be more taxing on the server than doing a CONCAT of all the fields then doing a single LIKE. If you're doing LIKE %search% where it can appear anywhere in any of the fields, then CONCAT + LIKE would work just as good as LIKE OR LIKE OR ... results-wise. I don't know if it's faster/less intensive or not though. You'd have to do some tests. Also, comparisons like = should (if I recall) be faster than LIKE comparisons. So if you really meant to use =, do that instead. Let me know if any of that's unclear. I know I get some kooky ideas sometimes. Also, anyone see anything I screwed up or have thoughts on this matter? Good luck! -TG - Original Message - From: Jas [EMAIL PROTECTED] To: php-db@lists.php.net Date: Wed, 26 Sep 2007 12:08:53 -0600 Subject: [PHP-DB] Re: mysql statement [SOLVED] Got if figured out, needed a sub-select type of query: mysql SELECT * - FROM ( SELECT * FROM `orders` - WHERE `group` = groupname ) - AS orders UNION SELECT * FROM `orders` - WHERE `ordernum` LIKE 35132 - OR `price` LIKE 35132 - OR `partnum` LIKE 35132 - OR `vendor` LIKE 35132 - OR `purpose` LIKE 35132 - OR `tracking` LIKE 35132 - OR `contact` LIKE 35132 - AND `group` LIKE 'mac' - ORDER BY `ordernum` - LIMIT 0 , 30; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] mysql statement
I am looking for some advice on how to achieve something and so far have been unable to do what I am looking to do. Here is the query I am using: mysql SELECT * - FROM `orders` - WHERE `ordernum` LIKE 35132 - OR `price` LIKE 35132 - OR `partnum` LIKE 35132 - OR `vendor` LIKE 35132 - OR `purpose` LIKE 35132 - OR `tracking` LIKE 35132 - OR `contact` LIKE 35132 - AND `group` LIKE 'mac' - ORDER BY `ordernum` - LIMIT 0 , 30; First here is the table structure: mysql describe orders; +-+--+--+-+-++ | Field | Type | Null | Key | Default | Extra | +-+--+--+-+-++ | id | int(255) | NO | PRI | | auto_increment | | ordernum| int(10) | NO | | || | date| varchar(60) | NO | | || | time| varchar(20) | NO | | || | group | varchar(20) | NO | | || | quantity| int(10) | NO | | || | description | varchar(255) | NO | | || | price | decimal(3,0) | NO | | || | partnum | varchar(40) | NO | | || | vendor | varchar(65) | NO | | || | purpose | varchar(255) | NO | | || | tracking| varchar(120) | NO | | || | contact | varchar(255) | NO | | || | eta | varchar(50) | NO | | || | department | varchar(125) | NO | | || | notes | varchar(255) | NO | | || +-+--+--+-+-++ 16 rows in set (0.00 sec) I am trying to essentially LIMIT all records returned to be limited by the `group` field so I can search for records and limit the rows returned by that one field. Any tips? TIA. jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Data not fetching in the textfield.
You could always do something like:
while( $x = myql_fetch_array( $results ) ) {
list( $var1, $var2, $var3, ... ) = $x;
echo input name=storename type=text value=$var1;
}
Chris Carter wrote:
I am trying to fetch data through this code:
My code:
for($i=0;$imysql_num_rows($results);$i++)
{
$rows = mysql_fetch_array($results);
echo 'div' ;
echo 'div class=fieldrow' ;
echo ' ' ;
echo ' label for=storenameStore
name:/label' ;
echo ' ' ;
echo ' input name=storename type=text
maxlength=35 id=storename
class=regForm disabled=disabled value=$rows[0]/input' ;
echo '/div' ;
The value option in the textfield is not fetching the data, neither is it
throwing errors. However the same code works perfectly as suggested by one
of expert Nabble users:
Nabble code:
for($i=0;$imysql_num_rows($results);$i++)
{
$rows = mysql_fetch_array($results);
echo(input name=\input1\ type=\text\
value=$rows[0]);
}
Please help.
Chris
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[PHP-DB] Re: mysql statement [SOLVED]
Got if figured out, needed a sub-select type of query: mysql SELECT * - FROM ( SELECT * FROM `orders` - WHERE `group` = groupname ) - AS orders UNION SELECT * FROM `orders` - WHERE `ordernum` LIKE 35132 - OR `price` LIKE 35132 - OR `partnum` LIKE 35132 - OR `vendor` LIKE 35132 - OR `purpose` LIKE 35132 - OR `tracking` LIKE 35132 - OR `contact` LIKE 35132 - AND `group` LIKE 'mac' - ORDER BY `ordernum` - LIMIT 0 , 30; Jas wrote: I am looking for some advice on how to achieve something and so far have been unable to do what I am looking to do. Here is the query I am using: mysql SELECT * - FROM `orders` - WHERE `ordernum` LIKE 35132 - OR `price` LIKE 35132 - OR `partnum` LIKE 35132 - OR `vendor` LIKE 35132 - OR `purpose` LIKE 35132 - OR `tracking` LIKE 35132 - OR `contact` LIKE 35132 - AND `group` LIKE 'mac' - ORDER BY `ordernum` - LIMIT 0 , 30; First here is the table structure: mysql describe orders; +-+--+--+-+-++ | Field | Type | Null | Key | Default | Extra | +-+--+--+-+-++ | id | int(255) | NO | PRI | | auto_increment | | ordernum| int(10) | NO | | || | date| varchar(60) | NO | | || | time| varchar(20) | NO | | || | group | varchar(20) | NO | | || | quantity| int(10) | NO | | || | description | varchar(255) | NO | | || | price | decimal(3,0) | NO | | || | partnum | varchar(40) | NO | | || | vendor | varchar(65) | NO | | || | purpose | varchar(255) | NO | | || | tracking| varchar(120) | NO | | || | contact | varchar(255) | NO | | || | eta | varchar(50) | NO | | || | department | varchar(125) | NO | | || | notes | varchar(255) | NO | | || +-+--+--+-+-++ 16 rows in set (0.00 sec) I am trying to essentially LIMIT all records returned to be limited by the `group` field so I can search for records and limit the rows returned by that one field. Any tips? TIA. jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] multiple fields all unique?
For this table to create 3 unique keys I did the following, in case it
helps someone else out.
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | | ||
| mac | varchar(100) | | | ||
| ip | varchar(100) | | | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
UPDATE TABLE hosts ADD UNIQUE mac (mac);
UPDATE TABLE hosts ADD UNIQUE hostname (hostname);
UPDATE TABLE hosts ADD UNIQUE ip (ip);
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | UNI | ||
| mac | varchar(100) | | UNI | ||
| ip | varchar(100) | | UNI | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
Now I have used the following to check if duplicate records exist before
updating:
?php
// Try and update with posted fields form html form
$update = mysql_query(UPDATE hosts SET hostname='$_POST[hostname]',
mac='$_POST[mac]', ip='$_POST[ip]', vlan='$_POST[vlan]' WHERE
id='$_SESSION[id]',$db);
$rows = mysql_affected_rows();
// Check results of operation
if($rows == 0) {
echo No matching records found;
} else {
echo Matching records found; }
Hope this helps anyone else, and thanks for the tip on MySQL's UNIQUE
field, wish I would have known it sooner, I wouldn't be so pissed off
from frustration.
Jas
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Re: [PHP-DB] multiple fields all unique? [almost solved]
Jas wrote:
For this table to create 3 unique keys I did the following, in case it
helps someone else out.
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | | ||
| mac | varchar(100) | | | ||
| ip | varchar(100) | | | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
UPDATE TABLE hosts ADD UNIQUE mac (mac);
UPDATE TABLE hosts ADD UNIQUE hostname (hostname);
UPDATE TABLE hosts ADD UNIQUE ip (ip);
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | UNI | ||
| mac | varchar(100) | | UNI | ||
| ip | varchar(100) | | UNI | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
Now I have used the following to check if duplicate records exist before
updating:
?php
// Try and update with posted fields form html form
$update = mysql_query(UPDATE hosts SET hostname='$_POST[hostname]',
mac='$_POST[mac]', ip='$_POST[ip]', vlan='$_POST[vlan]' WHERE
id='$_SESSION[id]',$db);
$rows = mysql_affected_rows();
// Check results of operation
if($rows == 0) {
echo No matching records found;
} else {
echo Matching records found; }
Hope this helps anyone else, and thanks for the tip on MySQL's UNIQUE
field, wish I would have known it sooner, I wouldn't be so pissed off
from frustration.
Jas
Sorry, but now I have one more question about finding the exact field
for a duplicate entry...
for instance, say you change the mac and hostname and there is a record
in the database with the same mac string, how can I flag the field that
matched from the 3?
Thanks,
jas
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Re: [PHP-DB] multiple fields all unique? [almost solved]
If I do this statement:
mysql_query(UPDATE hosts SET hostname=\$_POST[hostname]\,
mac=\$_POST[mac]\, ip=\$_POST[ip]\, vlan=\$_POST[vlan]\ WHERE
id=\$_SESSION[id]\,$db)or die(mysql_error() . mysql_errno());
I get this error:
Duplicate entry '128.110.22.139' for key 41062
I have tried using these types of checks with no success:
$update = mysql_query(UPDATE hosts SET hostname=\$_POST[hostname]\,
mac=\$_POST[mac]\, ip=\$_POST[ip]\, vlan=\$_POST[vlan]\ WHERE
id=\$_SESSION[id]\,$db)or die(mysql_error() . mysql_errno());
$rows = mysql_affected_rows();
while($match = mysql_fetch_assoc($update)) {
echo $match[hostname];
echo $match[mac];
echo $match[ip]; }
if($rows == 0) {
echo update worked;
} else {
echo update didn't work; }
And...
while($match = mysql_fetch_object($update)) {
And..
while($match = mysql_fetch_array($update)) {
So far everything I have tried will not allow me to find the exact field
and contents of a record that matches an existing record in the
database. See below for details on database structure etc.
Any help is appreciated,
jas
Jas wrote:
Jas wrote:
For this table to create 3 unique keys I did the following, in case it
helps someone else out.
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | | ||
| mac | varchar(100) | | | ||
| ip | varchar(100) | | | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
UPDATE TABLE hosts ADD UNIQUE mac (mac);
UPDATE TABLE hosts ADD UNIQUE hostname (hostname);
UPDATE TABLE hosts ADD UNIQUE ip (ip);
+--+--+--+-+-++
| Field| Type | Null | Key | Default | Extra |
+--+--+--+-+-++
| id | int(11) | | PRI | NULL| auto_increment |
| hostname | varchar(100) | | UNI | ||
| mac | varchar(100) | | UNI | ||
| ip | varchar(100) | | UNI | ||
| vlan | varchar(100) | | | ||
+--+--+--+-+-++
Now I have used the following to check if duplicate records exist
before updating:
?php
// Try and update with posted fields form html form
$update = mysql_query(UPDATE hosts SET hostname='$_POST[hostname]',
mac='$_POST[mac]', ip='$_POST[ip]', vlan='$_POST[vlan]' WHERE
id='$_SESSION[id]',$db);
$rows = mysql_affected_rows();
// Check results of operation
if($rows == 0) {
echo No matching records found;
} else {
echo Matching records found; }
Hope this helps anyone else, and thanks for the tip on MySQL's UNIQUE
field, wish I would have known it sooner, I wouldn't be so pissed off
from frustration.
Jas
Sorry, but now I have one more question about finding the exact field
for a duplicate entry...
for instance, say you change the mac and hostname and there is a record
in the database with the same mac string, how can I flag the field that
matched from the 3?
Thanks,
jas
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Re: [PHP-DB] multiple fields all unique? [almost solved]
John W. Holmes wrote: [snip] When you update the table with an existing mac value, the error will be similar to Duplicate value for Key XX where XX is what key was duplicated. I can't remember if the keys start at zero or one, but your ID column will be the first key, then mac, hostname, and finally ip (in the order they were created). So, if you examine the result of mysql_error() and it say duplicate for key 2, then the mac column was duplicated. Although this sounds a little harder than doing a SELECT prior to and just comparing values, it lets you do this with just a single query and only have extra processing upon errors instead of every single update. [/snip] Your right, I am able to see which record matches (not the record I am updating) I think you just answered my question by stating that Although this sounds a little harder than doing a SELECT prior to and just comparing values, it lets you do this with just a single query and only have extra processing upon errors instead of every single update. I suppose that is what I am going to have to do before doing the update. If you know of a different method please elaborate. thanks again, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] multiple fields all unique? [almost solved]
[snip]
You're not going to be able to fetch anything from the result set because
you're excuting an UPDATE query, not a SELECT.
You also do not want to die() when the query fails, otherwise you won't be
able to react to the error. Execute the query, then check mysql_error() for
a value. If it contains a value, then the query failed more than likely
because of a duplicate key.
[snip]
Well looks like I need to brush up on my php - mysql functions. =)
For anyone that has been following this thread here is a simple way to
catch the duplicate and show the user the contents:
?php
$update = mysql_query(UPDATE hosts SET hostname=\$_POST[hostname]\,
mac=\$_POST[mac]\, ip=\$_POST[ip]\, vlan=\$_POST[vlan]\ WHERE
id=\$_SESSION[id]\,$db);
$rows = mysql_affected_rows();
if([EMAIL PROTECTED]($update)) {
echo No match found;
} else {
echo Match found;
$error = preg_match(/^[']$/,mysql_errno($update));
$sql = mysql_query(select * from hosts where id = $error)
while($x = mysql_fetch_array($sql)) {
list($id,$host,$mac,$ip,$vlan) = $x; }
}
?
Might want to check the preg_match string to look for everything between
' and ' but other than that you should be able to list the record that
your update statement is having a problem with.
HTH
Jas
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[PHP-DB] multiple fields all unique?
I am running into a problem and I have yet to find an elegant solution for it. I have three fields in a database that cannot have duplicates of these particular fields. +--+---++ | hostname | mac | ip | +--+---++ | pmac-1 | 00:30:65:6c:ea:cc | 128.110.22.15 | +--+---++ | pmac-2 | 00:30:65:6c:e0:cc | 128.110.22.16 | +--+---++ Now I have a simple form that opens up a record and allows users to modify 2 of the 3 fields, mac ip. In order for the user to save the record and overwrite the current contents of the database I first need to do a check to see if any of the 2 fields being updated match an existing record. For example say the user modifies the ip field from 128.110.22.15 to 128.110.22.16, because there is an entry that matches that change the record does not update, or vice versa for the mac fields. Has anyone every performed such a feat as to check for matching fields before updating? And if so could you show me some code that used to accomplish this. I have written my own but the if - else statements are getting ridiculous. Thanks, jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] multiple fields all unique?
John W. Holmes wrote: From: Jason Gerfen [EMAIL PROTECTED] Yeah, I have never used a unique field for the database, and you are right it is a mysql db. Now is a good time to do it the right way, then... ---John Holmes... Yes definately, I am trying to convert my exisiting fields to a unique type and am having no luck what-so-ever. This is the table format: +--+--+--+-+-++ | Field| Type | Null | Key | Default | Extra | +--+--+--+-+-++ | id | int(11) | | PRI | NULL| auto_increment | | hostname | varchar(100) | | | || | mac | varchar(100) | | | || | ip | varchar(100) | | | || | vlan | varchar(100) | | | || +--+--+--+-+-++ 5 rows in set (0.00 sec) and this is the command I am tying to run to modify the hostname, mac ip fields to unique. alter hosts add unique mac; etc. I have such a hard time with mysql's manual. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Checking for duplicate records before update?
Problem. I have a database table that looks like this:
eg.
+--+--+---+---+-+
| id | hostname | mac | ip| vlan|
+--+--+---+---+-+
| 1014 | MTPC-01 | 00:02:B3:A2:9D:ED | 155.97.15.11 | Vlan-29 |
| 1015 | MTPC-02 | 00:02:B3:A2:B6:F4 | 155.97.15.12 | Vlan-29 |
This table will hold a very large number of entries, all unique. I have
created a simple HTML form which updates the entries but before it does
it checks to see if records already exist that contain either the same
IP value, the same MAC value or the same Hostname value. If any of
these items exist it currently discards the posted informaiton and
prompts the user to re-enter because I cannot have duplicate entries of
either the hostname, mac, ip fields.
Here is my function:
eg.
$x = mysql_query(SELECT * FROM $table WHERE hostname =
'$_POST[hostname]' OR ip = '$_POST[ip]' OR mac = '$_POST[mac]' NOT id =
'$_SESSION[id01]')or die(mysql_error());
$num = mysql_num_rows($x);
if($num == 0) {
unset($_SESSION['search']);
require 'dbase.inc.php';
$table = hosts;
$sql = @mysql_query(UPDATE $table SET hostname =
\$_POST[hostname]\, mac = \$_POST[mac]\, ip = \$_POST[ip]\, vlan =
\$_POST[vlan]\ WHERE id = \$_SESSION[id01]\)or die(mysql_error());
echo Form worked!;
} elseif ($num != 0) {
unset($_SESSION['search']);
echo Form didn't work because 1 of the 3 fields were present in
another record!!! Please try again.;
} else {
echo RTM again please!; }
I think what I really need to know is if there is a quick way to sort
through the results of the current database records and do comparisons
against the form. If any one has done something like this before could
you please show me an example or point me to a good resource. Thanks in
advance.
Jas
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[PHP-DB] sql errors?
I have checked and rechecked my code on these php to mysql statements and for some reason I either get no data (through php, where through a command line I get all the data I need) or errors on the update function. Could someone tell me what I am doing wrong? $sql = mysql_query(SELECT hostname FROM $table WHERE vlan = $_POST[dhcp_hosts],$db)or die(mysql_error()); from the command line SELECT hostname FROM hosts WHERE vlan = 'Vlan-22'; And I get results... I don't get it. And for the update statement I am having problems with... $sql_global = mysql_query(UPDATE $tble SET dn = '$_POST[dn]' SET lt = '$_POST[lt]' SET mlt = '$_POST[mlt]' SET msc01 = '$_POST[msc01]' SET msc02 = '$_POST[msc02]' SET msc03 = '$_POST[msc03]' SET msc04 = '$_POST[msc04]' WHERE id = '1',$db)or die(mysql_error()); If you echo the posted vars they are all present and the same statement works in other places of the script and from the command line. Any help is appreciated. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] form not working...
For some reason my form to edit selected items is not working, I think I
need a new set of eyes.
Any help is appreciated...
Jas
// inc.php
function db_retrieve() {
require 'dbase.inc.php';
$tble = $_SESSION['table'];
$show = @mysql_query(SELECT * FROM $tble LIMIT 0, 15,$db)or
die(mysql_error());
while ($row = @mysql_fetch_array($show)) {
@list($nItemSKU, $bActive, $nUserID, $nItemConfig, $tItemType,
$tDepartment, $blSerialNumber, $nPropertyNumber, $blHostName, $nIPID,
$tRoomNumber, $tNotes) = $row;
$_SESSION['table'] .= tr class=fntMAINtd valign=top
align=left$nItemSKU/td
td valign=top align=left$bActive/td
td valign=top align=left$nUserID/td
td valign=top align=left$nItemConfig/td
td valign=top align=left$tItemType/td
td valign=top align=left$tDepartment/td
td valign=top align=left$blSerialNumber/td
td valign=top align=left$nPropertyNumber/td
td valign=top align=left$blHostName/td
td valign=top align=left$nIPID/td
td valign=top align=left$tRoomNumber/td
td valign=top align=left$tNotes/td
td valign=top align=leftinput name=edit type=checkbox
value=$blSerialNumber/td
/tr; }
$_SESSION['number'] = mysql_num_rows($show); }
?php
session_start();
require 'scripts/inc.php';
if ((!isset($_POST['edit'])) or ($_POST = )) {
$_SESSION['table'] = t_items;
call_user_func(db_retrieve);
} elseif ((isset($_POST['edit'])) or (!$_POST = )) {
require 'scripts/dbase.inc.php';
$tble = $_SESSION['table'];
$edt = mysql_query(SELECT * FROM $tble WHERE
blSerialNumber=\$edit\,$db)or die(mysql_error());
while ($row = mysql_fetch_array($edt)) {
list($nItemSKU, $bActive, $nUserID, $nItemConfig, $tItemType,
$tDepartment, $blSerialNumber, $nPropertyNumber, $blHostName, $nIPID,
$tRoomNumber, $tNotes) = $row;
$_SESSION['table'] .= tr class=fntMAIN
td valign=top align=leftinput name=nItemSKU type=text size=30
value=$nItemSKU/td
td valign=top align=leftinput name=bActive type=text size=30
value=$bActive$bActive/td
td valign=top align=leftinput name=nUserID type=text size=30
value=$nUserID$nUserID/td
td valign=top align=leftinput name=nItemConfig type=text
size=30 value=$nItemConfig$nItemConfig/td
td valign=top align=leftinput name=tItemType type=text size=30
value=$tItemType$tItemType/td
td valign=top align=leftinput name=tDepartment type=text
size=30 value=$tDepartment$tDepartment/td
td valign=top align=leftinput name=blSerialNumber type=text
size=30 value=$blSerialNumber$blSerialNumber/td
td valign=top align=leftinput name=nPropertyNumber type=text
size=30 value=$nPropertyNumber$nPropertyNumber/td
td valign=top align=leftinput name=blHostName type=text
size=30 value=$blHostName$blHostName/td
td valign=top align=leftinput name=nIPID type=text size=30
value=$nIPID$nIPID/td
td valign=top align=leftinput name=tRoomNumber type=text
size=30 value=$tRoomNumber$tRoomNumber/td
td valign=top align=leftinput name=tNotes type=text size=30
value=$tNotes$tNotes/td
td valign=top align=leftinput name=save type=button value=Save
class=frmBTNS/td
/tr; }
}
?
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
?php echo $_SESSION['table']; ?
input name=edit type=button value=edit selected items
class=frmBTNS
/form
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Re: [PHP-DB] form not working...
Yeah, sorry that code wasn't updated... after the $_session['table'] is
called the input button doesn't attempt to post the data form the
$_session['table'] if that makes sense... all the error checking isn't the
problem
jas
Jeffrey N Dyke [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
i'd be curious what your actual problem is?
here's a question though.,...would these ever work?.
snip
if ((!isset($_POST['edit'])) or ($_POST = )) {
$_SESSION['table'] = t_items;
call_user_func(db_retrieve);
} elseif ((isset($_POST['edit'])) or (!$_POST = )) {
/snip
You're basically setting a superglobal array equal to an empty string -
. this should at least be $_POST == ...but
instead, i'd use
if ((!isset($_POST['edit'])) or (empty($_POST)) {
--and --
} elseif ((isset($_POST['edit'])) or (!empty($_POST))) {
hth
jeff
jas
[EMAIL PROTECTED]To:
[EMAIL PROTECTED]
.comcc:
Subject: [PHP-DB] form not
working...
11/10/2003 12:38
PM
For some reason my form to edit selected items is not working, I think I
need a new set of eyes.
Any help is appreciated...
Jas
// inc.php
function db_retrieve() {
require 'dbase.inc.php';
$tble = $_SESSION['table'];
$show = @mysql_query(SELECT * FROM $tble LIMIT 0, 15,$db)or
die(mysql_error());
while ($row = @mysql_fetch_array($show)) {
@list($nItemSKU, $bActive, $nUserID, $nItemConfig, $tItemType,
$tDepartment, $blSerialNumber, $nPropertyNumber, $blHostName, $nIPID,
$tRoomNumber, $tNotes) = $row;
$_SESSION['table'] .= tr class=fntMAINtd valign=top
align=left$nItemSKU/td
td valign=top align=left$bActive/td
td valign=top align=left$nUserID/td
td valign=top align=left$nItemConfig/td
td valign=top align=left$tItemType/td
td valign=top align=left$tDepartment/td
td valign=top align=left$blSerialNumber/td
td valign=top align=left$nPropertyNumber/td
td valign=top align=left$blHostName/td
td valign=top align=left$nIPID/td
td valign=top align=left$tRoomNumber/td
td valign=top align=left$tNotes/td
td valign=top align=leftinput name=edit type=checkbox
value=$blSerialNumber/td
/tr; }
$_SESSION['number'] = mysql_num_rows($show); }
?php
session_start();
require 'scripts/inc.php';
if ((!isset($_POST['edit'])) or ($_POST = )) {
$_SESSION['table'] = t_items;
call_user_func(db_retrieve);
} elseif ((isset($_POST['edit'])) or (!$_POST = )) {
require 'scripts/dbase.inc.php';
$tble = $_SESSION['table'];
$edt = mysql_query(SELECT * FROM $tble WHERE
blSerialNumber=\$edit\,$db)or die(mysql_error());
while ($row = mysql_fetch_array($edt)) {
list($nItemSKU, $bActive, $nUserID, $nItemConfig, $tItemType,
$tDepartment, $blSerialNumber, $nPropertyNumber, $blHostName, $nIPID,
$tRoomNumber, $tNotes) = $row;
$_SESSION['table'] .= tr class=fntMAIN
td valign=top align=leftinput name=nItemSKU type=text
size=30
value=$nItemSKU/td
td valign=top align=leftinput name=bActive type=text size=30
value=$bActive$bActive/td
td valign=top align=leftinput name=nUserID type=text size=30
value=$nUserID$nUserID/td
td valign=top align=leftinput name=nItemConfig type=text
size=30 value=$nItemConfig$nItemConfig/td
td valign=top align=leftinput name=tItemType type=text
size=30
value=$tItemType$tItemType/td
td valign=top align=leftinput name=tDepartment type=text
size=30 value=$tDepartment$tDepartment/td
td valign=top align=leftinput name=blSerialNumber type=text
size=30 value=$blSerialNumber$blSerialNumber/td
td valign=top align=leftinput name=nPropertyNumber type=text
size=30 value=$nPropertyNumber$nPropertyNumber/td
td valign=top align=leftinput name=blHostName type=text
size=30 value=$blHostName$blHostName/td
td valign=top align=leftinput name=nIPID type=text size=30
value=$nIPID$nIPID/td
td valign=top align=leftinput name=tRoomNumber type=text
size=30 value=$tRoomNumber$tRoomNumber/td
td valign=top align=leftinput name=tNotes type=text size=30
value=$tNotes$tNotes/td
td valign=top align=leftinput name=save type=button
value=Save
class=frmBTNS/td
/tr; }
}
?
form action=?php echo $_SERVER['PHP_SELF']; ? method=post
?php echo $_SESSION['table']; ?
input name=edit type=button value=edit selected items
class=frmBTNS
/form
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[PHP-DB] Add items to field using update?
Not sure how to do this but I have a table which looks like: CREATE TABLE cm_sessions ( session varchar(255) NOT NULL default '', ipaddy varchar(30) NOT NULL default '', host varchar(255) NOT NULL default '', pages longtext NOT NULL, referrer varchar(255) NOT NULL default '', hck varchar(15) NOT NULL default '', nrml varchar(15) NOT NULL default '', hour varchar(15) NOT NULL default '', date_stamp varchar(255) NOT NULL default '', PRIMARY KEY (session), UNIQUE KEY session (session) ) TYPE=MyISAM; Now everytime a new page is reached I want to update only the 'pages' field, here is the current sql statement that overwrites the contents of the 'pages' field. UPDATE $table SET pages = '.$PHP_SELF.' WHERE session = '$holy_cow' any help on this would be great. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] SQL update help
Here is my statement UPDATE $table SET pages = $PHP_SELF WHERE session = $holy_cow However it will not update I recieve this error: You have an error in your SQL syntax near '/admin/login.done.php WHERE session = 5d44389bd70881131b9ea7573eba34d4' at line 1 Any help would be appreciated Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Whats the best way to update...
I need to know the best way to update a record based on something like the
session id being the same, except the other fields are not. Any examples of
this would help...
$table = cm_sessions;
$sql_insert_form = INSERT INTO $table
(session,ipaddy,host,referrer,hck,nrml,hour,date_stamp) VALUES
('$holy_cow','$ipaddy','$host','$domain_b','$hck','-','$hour','$date');
$result = mysql_query($sql_insert_form,$dbh) or die(MYSQL_ERROR());
or
$sql_insert = UPDATE $table WHERE session = $HTTP_SESSION_VARS['session']
or if the session vars are the same it would insert a new row? I hope I am
asking the question correctly. Any help, examples would be great!
TIA
Jas
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[PHP-DB] Re: Benefits of assigning query to variable
sure there are, you can call the query in multiple places, for example:
$sql = mysql_query(INSERT INTO $table_name WHERE db_field =
$search_string);
Then say you have a routine to check for missing for elements and you would
like to log failed vs. successfull attempts your code flow would be
something like:
if (!$search_string !$var02) {
$sql;
} else {
$sql; }
HTH
Jas
Brian Graham [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Are there any benefits to assigning a query to a variable?
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[PHP-DB] Column count error?
I am not sure how to resolve this type of error, any help is appreciated.
TIA
Jas
/* Error message */
Column count doesn't match value count at row 1
/* Code to query db for username and password */
require '/home/bignickel.net/scripts/admin/db.php';
$db_table = 'auth_users';
$sql = SELECT * from $db_table WHERE un = \$u_name\ AND pw =
password(\$p_word\);
$result = @mysql_query($sql,$dbh) or die('Cannot execute query, please try
again later or contact the system administrator by email at
[EMAIL PROTECTED]');
/* Loop through records for matching pair */
$num = mysql_numrows($result);
if ($num !=0) {
print You have a valid username and password combination;
} else {
header(Location: blank.php); }
/* Table structure of db */
CREATE TABLE auth_users (
user_id int(11) NOT NULL auto_increment,
f_name varchar(255) default NULL,
l_name varchar(255) default NULL,
email_addy varchar(255) default NULL,
un text,
pw text,
PRIMARY KEY (user_id)
) TYPE=MyISAM;
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[PHP-DB] Re: Not sure what I am doing wrong here?
No kidding huh? Thanks a ton.
David Robley [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
In article [EMAIL PROTECTED], [EMAIL PROTECTED]
says...
Ok, now I am not sure why I cannot get this output into a variable so I
can
place it into a database table field. Any help would be greatly
appreciated.
?php
/* Check client info and register */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR);
$host = gethostbyaddr($REMOTE_ADDR); }
/* Start looking up users IP vs. WHOIS database for logging */
$nipaddy = ;
function message($msg){
echo $msg;
flush();
}
function arin($ipaddy){
$server = whois.arin.net;
if (!$ipaddy = gethostbyname($ipaddy))
$msg .= Can't IP Whois without an IP address.;
else{
if (! $sock = fsockopen($server, 43, $num, $error, 20)){
unset($sock);
$msg .= Timed-out connecting to $server (port 43);
}
else{
fputs($sock, $ipaddy\n);
while (!feof($sock))
$buffer .= fgets($sock, 10240);
fclose($sock);
}
if (eregi(RIPE.NET, $buffer))
$nextServer = whois.ripe.net;
else if (eregi(whois.apnic.net, $buffer))
$nextServer = whois.apnic.net;
else if (eregi(nic.ad.jp, $buffer)){
$nextServer = whois.nic.ad.jp;
#/e suppresses Japanese character output from JPNIC
$extra = /e;
}
else if (eregi(whois.registro.br, $buffer))
$nextServer = whois.registro.br;
if($nextServer){
$buffer = ;
message(Deferred to specific whois server:
$nextServer...brbr);
if(! $sock = fsockopen($nextServer, 43, $num, $error, 10)){
unset($sock);
$msg .= Timed-out connecting to $nextServer (port 43);
Hmm, seems your message is too long for my newsreader to include the lot.
However, it seems that you may not be returning a value from the function
arin(), so whan you do
/* Place results into a var */
$whois = arin($ipaddy); // Right now if this is here it will show in two
areas of the screen
natuarally nothing is placed in $whois.
Cheers
--
David Robley
Temporary Kiwi!
Quod subigo farinam
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[PHP-DB] Not sure what I am doing wrong here?
Ok, now I am not sure why I cannot get this output into a variable so I can
place it into a database table field. Any help would be greatly
appreciated.
?php
/* Check client info and register */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR);
$host = gethostbyaddr($REMOTE_ADDR); }
/* Start looking up users IP vs. WHOIS database for logging */
$nipaddy = ;
function message($msg){
echo $msg;
flush();
}
function arin($ipaddy){
$server = whois.arin.net;
if (!$ipaddy = gethostbyname($ipaddy))
$msg .= Can't IP Whois without an IP address.;
else{
if (! $sock = fsockopen($server, 43, $num, $error, 20)){
unset($sock);
$msg .= Timed-out connecting to $server (port 43);
}
else{
fputs($sock, $ipaddy\n);
while (!feof($sock))
$buffer .= fgets($sock, 10240);
fclose($sock);
}
if (eregi(RIPE.NET, $buffer))
$nextServer = whois.ripe.net;
else if (eregi(whois.apnic.net, $buffer))
$nextServer = whois.apnic.net;
else if (eregi(nic.ad.jp, $buffer)){
$nextServer = whois.nic.ad.jp;
#/e suppresses Japanese character output from JPNIC
$extra = /e;
}
else if (eregi(whois.registro.br, $buffer))
$nextServer = whois.registro.br;
if($nextServer){
$buffer = ;
message(Deferred to specific whois server: $nextServer...brbr);
if(! $sock = fsockopen($nextServer, 43, $num, $error, 10)){
unset($sock);
$msg .= Timed-out connecting to $nextServer (port 43);
}
else{
fputs($sock, $ipaddy$extra\n);
while (!feof($sock))
$buffer .= fgets($sock, 10240);
fclose($sock);
}
}
$buffer = str_replace( , nbsp;, $buffer);
$msg .= nl2br($buffer);
}
message($msg);
}
/* Place results into a var */
$whois = arin($ipaddy); // Right now if this is here it will show in two
areas of the screen
/* Start putting the data into the database table */
$table = sessions;
$sql_insert = INSERT INTO $table (ipaddy,whois) VALUES
('$ipaddy','$whois');
$result = mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR()); // No error
but what gets entered for the var $whois is arin(66.56.34.32)
?
?php echo $ipaddy; ?
?php echo arin($ipaddy; ? // These results should be placed in $whois
which is empty
?php echo $whois; ? // This returns arin(65.44.33.23) which is wrong
Any help on this would be appreciated!
Jas
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[PHP-DB] never seen this before..
I am getting an error when inserting data into a mysql database table. It
is telling me that there is a duplicate entry in the database everytime it
gets 10 records inserted to the table, if I flush the table and try it again
it works fine. Is there some kind of limit on some tables? Here is the
error php outputs once the table gets 10 records: Duplicate entry '10' for
key 1
Here is the code...
require '/home/web/b/bignickel.net/scripts/admin/db.php';
$table = demo_sessions;
$sql_insert = INSERT INTO $table (ipaddy,date) VALUES
('$ipaddy','$date') or die('Could not insert client info');
$result = mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
Any help on this would be great
TIA
Jas
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[PHP-DB] Re: never seen this before..
Here is the db table structure...
CREATE TABLE demo_sessions (
id varchar(255) NOT NULL auto_increment,
ipaddy varchar(255) NOT NULL,
date varchar(255) NOT NULL,
PRIMARY KEY (id)
);
Let me know if this is correct, I have a feeling that the varchar is my
problem but I could be wrong.
TIA
Jas
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
TIA --
column in your table denotes your primary key? After that, how is it
incremented? Is it auto or do you have some other form of it. I think
your code is correct so the error seems to be how the data ID is
generated.
gl -- Seth
Jas wrote:
I am getting an error when inserting data into a mysql database table.
It
is telling me that there is a duplicate entry in the database everytime
it
gets 10 records inserted to the table, if I flush the table and try it
again
it works fine. Is there some kind of limit on some tables? Here is the
error php outputs once the table gets 10 records: Duplicate entry '10'
for
key 1
Here is the code...
require '/home/web/b/bignickel.net/scripts/admin/db.php';
$table = demo_sessions;
$sql_insert = INSERT INTO $table (ipaddy,date) VALUES
('$ipaddy','$date') or die('Could not insert client info');
$result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
Any help on this would be great
TIA
Jas
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[PHP-DB] resource id#2 - ????
Not sure how to over come this, the result of a database query keeps giving
me this:
?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */
session_start();
if (isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES ('$ipaddy','$referrer','$date_stamp');
$result = mysql_query($sql_insert,$dbh) or die(Couldn't execute
insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = mysql_fetch_array($record)) {
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = mysql_query($sql_content,$dbh) or die(Couldn't execute
query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_properties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
?php echo $main; ?
table width=75% border=1
?php echo $current; ? // my results should be here but instead I get
Resource ID #2 printed wtf?
/tablebr
?php echo $ipaddy; ?br
?php echo $referrer; ?br
?php echo $date_stamp; ?br
?php echo $sql_insert; ?br
?php echo $sql_content; ?br
?php echo $id; ?br
?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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Re: [PHP-DB] resource id#2 - ????
Thanks a ton, I was really looking right past that variable.
Jas
Paul Dubois [EMAIL PROTECTED] wrote in message
news:p05111763b92d3addd3b4@[192.168.0.33]...
At 11:42 -0600 6/12/02, Jas wrote:
Not sure how to over come this, the result of a database query keeps
giving
me this:
?php
/* Get Ip address, where they came from, and stamp the time */
if (getenv(HTTP_X_FORWARDED_FOR)){
$ipaddy = getenv(HTTP_X_FORWARDED_FOR);
} else {
$ipaddy = getenv(REMOTE_ADDR); }
$referrer = $HTTP_REFERER;
$date_stamp = date(Y-m-d H:i:s);
/* Start session, and check registered variables */
session_start();
if (isset($HTTP_SESSION_VARS['ipaddy']) ||
isset($HTTP_SESSION_VARS['referrer']) ||
isset($HTTP_SESSION_VARS['date_stamp'])) {
$main = Video clips stored in Database;
/* Insert client info to database table */
require('/path/to/connection/class/con.inc');
$table_sessions = dev_sessions;
$sql_insert = INSERT INTO $table_sessions
(ipaddy,referrer,date_stamp) VALUES
('$ipaddy','$referrer','$date_stamp');
$result = @mysql_query($sql_insert,$dbh) or die(Couldn't execute
insert to database!);
/* Pull video info from database table into an array */
$table_content = dev_videos;
$sql_content = @mysql_query(SELECT * FROM $table_content,$dbh);
while ($row = @mysql_fetch_array($record)) {
$record isn't defined anywhere. Do you mean $sql_content?
$id = $row['id'];
$file_name = $row['file_name'];
$file_size = $row['file_size'];
$file_properties = $row['file_properties'];
$file_codec = $row['file_codec'];
$file_author = $row['file_author'];
$result = @mysql_query($sql_content,$dbh) or die(Couldn't execute
query on database!);
/* loop through records and print results of array into table */
$current .=
trtd$id/tdtd$file_name/tdtd$file_size/tdtd$file_properties
/tdtd$file_codec/tdtd$file_author/td/tr; }
} else {
/* Start a new session and register variables */
session_start();
session_register('ipaddy');
session_register('referrer');
session_register('date_stamp'); }
?
Just to make sure everything is working correctly I have echoed every
statement to the screen so I may see what's going on like so:
?php echo $main; ?
table width=75% border=1
?php echo $current; ? // my results should be here but instead I get
Resource ID #2 printed wtf?
/tablebr
?php echo $ipaddy; ?br
?php echo $referrer; ?br
?php echo $date_stamp; ?br
?php echo $sql_insert; ?br
?php echo $sql_content; ?br
?php echo $id; ?br
?php echo $file_name; ?br
?php echo $file_size; ?br
Any help would be great!
Jas
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[PHP-DB] new pair of eyes
I think I am missing something: $db_name = database; $table_name = auth_users; $sql = DELETE user_id, f_name, l_name, email_addy, un, pw FROM $table_name WHERE user_id = \$user_id\, $dbh; $result = mysql_query($sql, $dbh) or die (Could not execute query. Please try again later.); Its not deleting the records based on a hidden field in a form named $user_id Any help would be great! Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: new pair of eyes
Sorry, got it working, just needed to remove user_id, f_name etc. and it works fine. Jas [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I think I am missing something: $db_name = database; $table_name = auth_users; $sql = DELETE user_id, f_name, l_name, email_addy, un, pw FROM $table_name WHERE user_id = \$user_id\, $dbh; $result = @mysql_query($sql, $dbh) or die (Could not execute query. Please try again later.); Its not deleting the records based on a hidden field in a form named $user_id Any help would be great! Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] how to pull array out?
Not sure how to do this, but I need to pull the contents of an array into
editable text fields... As of yet I have not found a way to accomplish
this:
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
while ($row = mysql_fetch_row($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
It only echoes the resource id, any help or examples would be great
Jas
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Re: [PHP-DB] how to pull array out?
Mr. Wong, You are no help. As you can probably see I am NOT an experienced php developer!! Posting to this list, reading the manual, and scouring sites for tutorials is all I have to provide me with a way to maybe begin doing writting php apps. I did read your reply about feeding the results into an array however I have not been able to accomplish what I have set out to do. A GOOD EXPLANATION IS ALL WE NEED!!! You posting stuff on how we should read the manual does not suffice! People post to this newsgroup to get answers to problems they cannot figure out. I am sure you have been working with php enough to know how to accomplish anything you need, and for the rest of us we rely on examples, etc to get it done. I appriciate you taking the time to reply to mine and others posts, however if you do not have anything nice to say please keep it to yourself. I am trying to find an answer to a problem and reading your RTFM does not help. Thannks again, Jas Jason Wong [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... On Friday 31 May 2002 00:19, Jas wrote: Not sure how to do this, but I need to pull the contents of an array into editable text fields... As of yet I have not found a way to accomplish this: Good grief, I already posted the answer yesterday. Don't you read the replies? Even if you don't, spending a little time googling for mysql php tutorial would get you numerous examples. $table = auth_users; $record = @mysql_query(SELECT * FROM $table WHERE user_id = '$user_id',$dbh); [snip] echo $record; It only echoes the resource id, any help or examples would be great I repeat, mysql_query() returns a resource-id which you need to feed into mysql_fetch_array() to get the actual record. -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* The chief danger in life is that you may take too many precautions. -- Alfred Adler */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: Re[2]: [PHP-DB] how to pull array out?
Ok, given there are people out there that expect people to do the work for
them, however if you look at my original post (today) you will see that I
have queried the database, feed the results into an array, then called the
array elements into my form, however simple this may be for you it is still
not pulling out the results of the array into my form. I do believe I am
missing something, and doing homework is a part of my everyday routine
trying to develop web based apps. The code again:
$table = auth_users; // names the table to query correct?
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh); // pulls all records in relation to $user_id correct?
while ($row = mysql_fetch_array($record)) { // pulls results of query
into an array correct?
$user_id = $row['user_id']; // assigns unique names for my different
editable regions correct?
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= form name=\$user_id\ method=\post\
action=\del_account.php\
bEdit Account $user_id/bbr // echoes the $user_id that is going
to be edited correct?
First Name:input type=\text\ name=\f_name\ size=\30\
maxlength=\30\ value=\$f_name\br // displays text box to edit with
$f_name variable from resulting array correct?
Last Name:input type=\text\ name=\l_name\ size=\30\
maxlength=\30\ value=\$l_name\br // displays text box to edit with
$l_name variable from resulting array correct?
Email:input type=\text\ name=\email_addy\ size=\30\
maxlength=\30\ value=\$email_addy\br // displays text box to edit
with $email_addy variable from resulting array correct?
User Name:input type=\text\ name=\un\ size=\30\ maxlength=\30\
value=\$un\br // displays text box to edit with $un variable from
resulting array correct?
Password:input type=\password\ name=\pw\ size=\30\
maxlength=\30\br // displays text box to edit with $pw variable from
resulting array correct?
Confirm Password:input type=\password\ name=\cpw\ size=\30\
maxlength=\30\
nbsp;input type=\submit\ name=\add\ value=\edit
user\nbsp;nbsp;input type=\reset\ name=\reset\ value=\reset\;
Any help or examples are appreciated. Please understand I am trying to
learn this scripting language and RTFM does not help as most of the time I
post here as a LAST resort, i.e. after I have indeed RTFM'ed.
Thanks again,
Jas
Julie Meloni [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Jas (and anyone else) -
With all due respect, you're acting like a troll. Posting a question,
getting MANY correct answers, then reposting the question and bitching
about not understanding the answers, well, that's troll-like.
Granted, many responses on this list over the past while have been
sarcastic, but usually also have the answer in them. The tone comes
from people asking the questions not doing their own homework.
e.g. http://www.tuxedo.org/~esr/faqs/smart-questions.html
Mailing lists such as this one are not intended to give people
verbatim answers to their problems, such as I have to write an
application for my university class, please do it for me and the
like. Instead, these lists will help those who attempt to help
themselves.
The fact of the matter is, RTFM is the right answer. The very
simplistic method of reading the results of a MySQL query are right
there in the manual, in the MySQL query functions section. Also,
they're in zillions of tutorials, any basic book on PHP, and so on.
In short:
1) connect to db server -- mysql_connect()
2) select db -- mysql_select_db()
3) issue query -- mysql_query()
4) from there you get a result identifier. You feed that into
mysql_result() or mysql_fetch_array() or mysql_fetch_row()
Read the manual for the differences -- including examples.
You say:
J rest of us we rely on examples, etc to get it done.
They ARE in the manual, which is pretty much the best manual out there
for just about anything.
Good luck,
- Julie
-- Julie Meloni
-- [EMAIL PROTECTED]
-- www.thickbook.com
Find Sams Teach Yourself MySQL in 24 Hours at
http://www.amazon.com/exec/obidos/ASIN/0672323494/thickbookcom-20
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Re: Re[2]: [PHP-DB] how to pull array out?
SUCCESS
Ok, I have fixed the problem. Here is the code, I hope it helps anyone that
comes accross something like this. Sorry for being a troll.
Jas
// Displays select form of current users in database table for editing
$table = users;
$record = @mysql_query(SELECT * FROM $table,$dbh);
while ($row = mysql_fetch_array($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw'];
$option .=OPTION VALUE=\$user_id\$f_name $l_name/OPTION;
$var_form = pFORM METHOD=\post\ ACTION=\edit_account.php\
bPlease select the user you would like to edit from the list/bbr
SELECT NAME=\blank\$option/SELECT
INPUT TYPE=\submit\ NAME=\submit\ VALUE=\select\
/FORMbr;
// Takes value of blank and looks for a record matching selected user and
puts the results into a form for editing
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$blank',$dbh);
while ($row = mysql_fetch_array($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\$f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\$l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\$email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\$un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\$pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\$pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
*** The problem was that I was looking for the variable $user_id in my
select statement not the option name which is blank (just me getting ahead
of myself, sorry once again) ***
Natalie Leotta [EMAIL PROTECTED] wrote in message
7546CB15C0A1D311BF0D0004AC4C4B0C024AC468@SSIMSEXCHNG">news:7546CB15C0A1D311BF0D0004AC4C4B0C024AC468@SSIMSEXCHNG...
From what I've seen on the PHP.net site, it doesn't look like people use
the
@ in @mysql_query, and I know that I don't use it for sybase_query. Where
did it come from?
Assuming that's not the problem, I'd try printing out everything. Right
after your while, print $row - it should say array, then try printing
$row[0] and see if that works. Then try printing $row['user_id'].
You may also want to try doing a SELECT with the fieldnames, instead of
just
the *. Also, make sure you're typing the field names correctly - they are
case sensitive.
Hopefully one of these suggestions will help you.
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, May 30, 2002 2:18 PM
To: [EMAIL PROTECTED]
Subject: Re: Re[2]: [PHP-DB] how to pull array out?
Ok, given there are people out there that expect people to do the work
for
them, however if you look at my original post (today) you will see that I
have queried the database, feed the results into an array, then called the
array elements into my form, however simple this may be for you it is
still
not pulling out the results of the array into my form. I do believe I am
missing something, and doing homework is a part of my everyday routine
trying to develop web based apps. The code again:
$table = auth_users; // names the table to query correct?
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh); // pulls all records in relation to $user_id correct?
while ($row = mysql_fetch_array($record)) { // pulls results of query
into an array correct?
$user_id = $row['user_id']; // assigns unique names for my different
editable regions correct?
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= form name=\$user_id\ method=\post\
action=\del_account.php\
bEdit Account $user_id/bbr // echoes the
[PHP-DB] Resource ID#2 ????
Ok here is my problem, I set this up so a user selects a name form a select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2. Not
sure how I can over come this...
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\$f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\$l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\$email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\$un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\$pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\$pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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Re: [PHP-DB] Resource ID#2 ????
If you look at the previously posted code at the bottom of the form there is a echo for the sql select statement that is echoing Resource id #2 on the page. Now that error is the correct field id number in the database, I am just not sure how to itemize the data from that table, at least I think. Any help would be great! Jas Jason Wong [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... On Thursday 30 May 2002 00:17, Jas wrote: Ok here is my problem, I set this up so a user selects a name form a select box and that name or $user_id is then passed to this page so the user can edit the contact info etc. However it does not pull the selected $user_id and place each field into my form boxes, all I get is a Resource ID #2. Not sure how I can over come this... $table = auth_users; $record = @mysql_query(SELECT * FROM $table WHERE user_id = '$user_id',$dbh); Make liberal use of error checking (see manual) and mysql_error(). -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* grep me no patterns and I'll tell you no lines. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource ID#2 ????
How can I get the form below to list the single record in a db that matches
the request of $user_id?
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
while ($row = mysql_fetch_row($record)) {
$user_id = $row['user_id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$email_addy = $row['email_addy'];
$un = $row['un'];
$pw = $row['pw']; }
$var_form .= table width=\100%\ border=\0\ cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\un\ size=\30\ maxlength=\30\ value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
Thanks in advance,
Jas
Natalie Leotta [EMAIL PROTECTED] wrote in message
7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG">news:7546CB15C0A1D311BF0D0004AC4C4B0C024AC420@SSIMSEXCHNG...
I could be missing something, but it looks like you are using the result
of
the mysql_query as the actual result. It actually returns some weird
identifier. To access the real info you'd have to use something like
mysql_fetch_array to get it.
Check out this and see if it helps:
http://www.php.net/manual/en/function.mysql-query.php
-Natalie
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, May 29, 2002 12:18 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Resource ID#2
Ok here is my problem, I set this up so a user selects a name form a
select
box and that name or $user_id is then passed to this page so the user can
edit the contact info etc. However it does not pull the selected $user_id
and place each field into my form boxes, all I get is a Resource ID #2.
Not
sure how I can over come this...
$table = auth_users;
$record = @mysql_query(SELECT * FROM $table WHERE user_id =
'$user_id',$dbh);
$var_form .= table width=\100%\ border=\0\
cellpadding=\7\form
name=\$user_id\ method=\post\ action=\del_account.php\
trtd width=\20%\ colspan=\2\bEdit Account
$user_id/b/td/tr
trtd width=\20%\First Name:/tdtd width=\80%\input
type=\text\ name=\$f_name\ size=\30\ maxlength=\30\
value=\$f_name\font class=\copyright\i.e. John/font/td/tr
trtd width=\20%\Last Name:/tdtd width=\80%\input
type=\text\ name=\$l_name\ size=\30\ maxlength=\30\
value=\$l_name\font class=\copyright\i.e. Doe/font/td/tr
trtd width=\20%\Email:/tdtd width=\80%\input
type=\text\ name=\$email_addy\ size=\30\ maxlength=\30\
value=\$email_addy\font class=\copyright\i.e.
[EMAIL PROTECTED]/font/td/tr
trtd width=\20%\User Name:/tdtd width=\80%\input
type=\text\ name=\$un\ size=\30\ maxlength=\30\
value=\$un\font
class=\copyright\i.e. j-doe/font/td/tr
trtd width=\20%\Password:/tdtd width=\80%\input
type=\password\ name=\$pw\ size=\30\ maxlength=\30\font
class=\copyright\(password must be alpha-numeric, i.e.
pAs5w0rd)/font/td/tr
trtd width=\20%\Confirm Password:/tdtd
width=\80%\input type=\password\ name=\$pw\ size=\30\
maxlength=\30\font class=\copyright\please confirm password
entered/font/td/tr
trtd width=\20%\nbsp;/tdtd width=\80%\input
type=\submit\ name=\add\ value=\edit user\nbsp;nbsp;input
type=\reset\ name=\reset\ value=\reset\/td/tr
/form/table;
echo $record;
} else {
blah blah
}
Thanks in advance,
Jas
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[PHP-DB] Error? new set of eyes...
Hello all,
I have an error and I will be darned if I know why. Here is the code:
?php
require '/path/to/database/connection/class/db.php';
$table = portfolio;
$portfolio = mysql_query(SELECT * FROM $table,$dbh);
while ($sections = mysql_fetch_array($portfolio)) {
list($id, $2d, $3d, $web, $prog, $tut, $proj) = $sections; // I think the
error is on this line.
}
?
Then I just echo the contents of the database array like so.
?php echo $2d; ? // this is line 69
And this is the error I am recieving:
Parse error: parse error, expecting `T_VARIABLE' or `'$'' in
/localhost/portfolio.php on line 169
Any help would be great!!
Jas
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[PHP-DB] How to store results into a session variable..
I need a tutorial or example on how to take a result from an mysql query and place it into a session variable. Please help? Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] mysql results to a session variable?
I am just wondering if there is a way to pull the results from an mysql query into an session variable that can be passed from page to page. If there is a way to do this could someone please point out a good tutorial on this specific function or maybe an example of how to accomplish this. Any help would be great. thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Unique Field for text field...
Here is my problem:
?php
/* Connecting, selecting database */
$link = mysql_connect(localhost, , )
or die(Could not connect);
mysql_select_db(shopping) or die(Could not select database);
/* Performing SQL query */
$query = SELECT * FROM cart;
$result = mysql_query($query) or die(Query failed);
/* Printing results in HTML */
print table width=\100%\ border=\0\ cellspacing=\0\
cellpadding=\3\\ntr
td colspan=\6\bSelect item(s) for
purchase/b/td
/trtr
td width=\3%\
bgcolor=\#e4e4e4\uID/u/td
td width=\10%\
bgcolor=\#e4e4e4\uName/u/td
td width=\15%\
bgcolor=\#e4e4e4\uItem Number/u/td
td width=\18%\
bgcolor=\#e4e4e4\uDescription/u/td
td width=\3%\
bgcolor=\#e4e4e4\uCost/u/td
td width=\3%\ bgcolor=\#e4e4e4\uSale/u/td
td width=\8%\ bgcolor=\#e4e4e4\uAdd Item/u/td
/tr;
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
print \ttr\n;
foreach ($line as $col_value) {
print \t\ttd$col_value/td\n;
}
print td width=\10%\ form name=\add_item\ method=\post\
action=\add_item\
/form
input type=\text\ name=\add_num\
size=\3\ maxlength=\6\
input type=\submit\ name=\add\
value=\add\/form
/td\t/tr\n;
}
print /table\n;
/* Free resultset */
mysql_free_result($result);
/* Closing connection */
mysql_close($link);
?
I use this to query the database and put the results into a form for later
use... The problem I am having is how would I assign a unique field name for
each entry displayed? I am not sure how I can do this with the $col_value
simply echoing the results onto the page, if anyone could shed some light on
this I would appreciate it. Thanks in advance,
Jas
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[PHP-DB] If statement for sql statement
Ok here is my question... I have a form that reads the contents of a
directory, places the resulting files into a select box within a form for
further processing, code is as follows:
?php
$dir_name = /virtual/path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\OPTION VALUE=\--\
NAME=\--\--/OPTION;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
This portion works great... now on index_done.php3 it takes the resulting
selection and places it into the db table, code is as follows:
?php
$file_var = http://localhost/images/;;
$db_name = database;
$table_name = image_path;
$connection = mysql_connect(localhost, user_name, password) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET image01_t=\$file_var$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
What I need to know is if there is a way to create an if statement that
would either update or not update a selected field in the database. Below
is a piece I would assume would work however I am not sure of the mysql
command to not update anything. Any help would be great.
Jas
if( $file_name = '--' ) $sql = dont update table;
else $sql = UPDATE $table_name SET image01_t = \$files\;
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[PHP-DB] Stuck on db entry from select box...
Ok here is my problem and where I am at this far, first thing I needed to do
is to read the contents of a directory and place the results into a select
box in a form (accomplished), now where I am stuck is passing the selection
from said select box to another script so it may be put into a mysql
database table... my code is as follows:
?php
$dir_name = /path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = database_name;
$table_name = cm_index;
$connection = mysql_connect(localhost, username, password) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET file_name=\$file_name\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
I think I need a new pair of eyes to review this and let me know where I am
going wrong, also on another note I don't want to put the file in the
database table I want to put the path to the file in the database table, any
help would be greatly appriciated. =)
Jas
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Re: [PHP-DB] Stuck on db entry from select box...
Ok I made the changes you suggested and unfortunately I am getting the same
results... Could not execute query, I manually put entries into the tables
so that the update command would just overwrite the current contents. Here
is my code this far, I think my problem (correct me if I am wrong) lies in
the fact that each of select boxes does not contain a unique name, needless
to say I am stuck:
?php
$dir_name = /path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
Select boxes are filled with files from specified folder... then echoed to
screen
? echo $file_list; ?
Working fine... when user click on image using said select box it links to
this file, code is as follows:
?php
$db_name = db_name;
$table_name = cm_index;
$connection = mysql_connect(localhost, user_name, password) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET file_name=\$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
Still getting could not execute query... sorry about not knowing where I am
going wrong but still new to these advanced functions. Thanks in advance,
Jas
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Re: [PHP-DB] Stuck on db entry from select box...
Upon doing a print $sql before the mysql_querie I get this output; UPDATE cm_index SET ad01_t=$filesCould not execute query. Please try again later -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] how to question...
Ok here is what I would like to accomplish, I am sure someone has done something like this but I need an example to go from since I am still not able to write php from scratch... I need to write a function to count the contents of a directory, like the file names and place them into an array so I can use a select box to select the file path in which I would like to save to database, like I said I don't want someone to do it for me, I just need a resource such as a tutorial or example to run with. Thanks a ton, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] calculations based on checkbox value
Ok here is my problem, currently I have a form that calculates the number of words in a given textarea. Now what I need to accomplish is using a checkbox to subtract a value based on said checkbox. I have looked for a tutorial on such an item and did not find one so I am asking if anyone out there has accomplished a similar function and if so could you point out a good tutorial or let me see some sample code? Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
Yeah that didn't work either... If I put the same value (
value=\{$record['wel_area']}\ ) on a text line it works just fine however
using the same value for a textarea doesnt give me any results nor does it
give me an error message to go on, if there is a way to output the code to
the screen like a dos echo command that would help out alot. Thanks in
advance,
Jas
Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Try this:
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is it
displays the textbox but not the results of the query. form
name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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Re: [PHP-DB] textarea and mysql query
Sorry, the connection is established already so that part works just fine.
It is getting the results of said connection \ query to appear in a textarea
vs. a text line.
Jason Wong [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
On Monday 18 March 2002 14:53, jas wrote:
I need some help with a query to place the results within a text area...
here is the code and so far every time I try it out all that happens is
it
displays the textbox but not the results of the query.
form name=wel_area method=post action=index_confirm.php3
target=box ?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh));
echo textarea name='wel_area' value='{$record['wel_area']}' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n;
?
/form
If this is *all* your code then you're missing the important step of
connecting to the mysql server and selecting a database.
Have a look at the examples in the manual.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
/*
Most people don't need a great deal of love nearly so much as they need
a steady supply.
*/
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Re: [PHP-DB] textarea and mysql query
Yeah it is not multidimensional at all, a simple text blob is what should be
pulled into the textarea. I am not trying to query multipled record fields
at all. One field per textarea. Do you know of a good tutorial on pulling
sql queries into textareas? If you do please let me know as I have searched
php.net and other sites for a tutorial on textarea's but have not found any
nor is there any reference to such a problem in any of the php or mysql
books i have.
Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
What is the result of $record['wel_area']. If this is a multidimensional
array then you will need to access it differently...
Example:
echo $record[0][0].\n;
Try this is your text area and see if that comes out...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 18, 2002 12:07 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] textarea and mysql query
Yeah that didn't work either... If I put the same value (
value=\{$record['wel_area']}\ ) on a text line it works just fine
however
using the same value for a textarea doesnt give me any results nor does it
give me an error message to go on, if there is a way to output the code to
the screen like a dos echo command that would help out alot. Thanks in
advance, Jas Ray Hunter [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Try this:
form name=wel_area method=post action=index_confirm.php3
target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='.$record['wel_area'].' cols='25'
rows='8'/textareabrinput type='submit' name='save'
value='save'br\n; ?
/form
If not then let me know...
Thank you,
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Sunday, March 17, 2002 11:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] textarea and mysql query
I need some help with a query to place the results within a text
area... here is the code and so far every time I try it out all that
happens is it displays the textbox but not the results of the query.
form
name=wel_area
method=post action=index_confirm.php3 target=box
?php
$record = mysql_fetch_array(@mysql_query(SELECT wel_area FROM
cm_index,$dbh)); echo textarea name='wel_area'
value='{$record['wel_area']}' cols='25' rows='8'/textareabrinput
type='submit' name='save' value='save'br\n; ?
/form
Any help would be great.
Jas
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[PHP-DB] text area
I am wondering where I could find a good tutorial on displaying the contents of a database field within a textarea for editing. Any help would be great! Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Monitor files
I am wondering if there is a way to monitor the contents of a directory for changes and if a change is made, to record the name of the file changed the date etc and send it to a database and also an email... has anyone every seen anything like this and if so are there any intrustion detection open source s/w that is commercially available for download etc. Maybe a tutorial on the subject would be great. Thanks Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] records into an editable box
How would I change this sql statement to pull the db table and display it within an editable box within a form? Any help or tutorials would be great. ?php $result = mysql_query(SELECT $table FROM $field,$dbh) or die(Could not execute query, please try again later); echo Btext:/Bbr\n; printf(mysql_result($result,wel_area)); ? Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] records into an editable box
No dan I mean textbox... but I am assuming you dont have any idea where a good tutorial would be then. Thanks anyways. Jas Dan Brunner [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hello!! Do you mean a textarea?? And you want to populate the textarea with the data, right? Dan On Wednesday, March 6, 2002, at 01:34 AM, [EMAIL PROTECTED] wrote: How would I change this sql statement to pull the db table and display it within an editable box within a form? Any help or tutorials would be great. ?php $result = mysql_query(SELECT $table FROM $field,$dbh) or die(Could not execute query, please try again later); echo Btext:/Bbr\n; printf(mysql_result($result,wel_area)); ? Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] security
how can you find out what the php.ini is looking like? is there a way to
use php to get that info. i have used phpinfo() but i cannot see whether or
not file_uploads is disabled
Jas
Paul Burney [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED];
on 3/3/02 7:39 PM, Ric Mañalac at [EMAIL PROTECTED] appended
the following bits to my mbox:
i personally think that the developer still has
the control in making his php code secure. but how do you
think will this news affect php as one of the most popular
choice for web developers?
Probably doesn't belong so much on the PHP-DB list, since databases not
involved, but since some of you on the list may not be aware
In most cases, PHP security can be controlled by the developer, but *not* in
this case.
Basically, most php security problems stem from someone not properly
checking input and being sloppy when connecting to databases, etc.
This case, however, is an actual problem in the PHP server code, not
anything you would write. To summarize, if you have file_uploads enabled on
the server, php parses multipart/form-data data that is sent to the
script.
It does this for *any* file, not just the ones that have file uploads in
them. The bug is in that code and can be used by malicious parties to do
evil things on your server. It can be used against you even if you only
have one page on your server parsed by PHP and the hacker can find it.
The original report is here:
http://security.e-matters.de/advisories/012002.html
Basically you have three options:
1) Disable file_uploads, if you're not using them, in the php.ini file.
This works for PHP 4.0.3 or greater.
2) Apply the source patch to your source tree and rebuild. Works for PHP
3.0.18, 4.06, 4.1.0 and 4.1.1.
3) Upgrade to PHP 4.1.2
You should really do this as soon as possible. I'm sure someone will make a
Code Red type of infestation soon to exploit this bug soon. Evidently,
there is a crude exploit circulating.
Hope that helps.
Paul
?php
while ($self != asleep) {
$sheep_count++;
}
?
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[PHP-DB] Re: why won't session_start() work?
Ryan,
Do you have anything else between your ?php and ?. Need more code to
determine why.
Jas
Ryan Snow [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED];
when I do:
session_start();
session_register('signor');
session_register('username');
I get:
Warning: Cannot send session cookie - headers already sent by (output
started at /var/www/html/index.php:3) in
/var/www/html/index.php on line 14
Warning: Cannot send session cache limiter - headers already sent (output
started at /var/www/html/index.php:3) in
/var/www/html/index.php on line 14
Anybody know why?
Thanks.
Ry
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[PHP-DB] Re: why won't session_start() work?
Not to mention that you are registering two different variables... try
combining the two by putting a coma between them. i.e.
session_register('signor','username');
HTH
Jas
Jas [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED];
Ryan,
Do you have anything else between your ?php and ?. Need more code
to
determine why.
Jas
Ryan Snow [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED];
when I do:
session_start();
session_register('signor');
session_register('username');
I get:
Warning: Cannot send session cookie - headers already sent by (output
started at /var/www/html/index.php:3) in
/var/www/html/index.php on line 14
Warning: Cannot send session cache limiter - headers already sent
(output
started at /var/www/html/index.php:3) in
/var/www/html/index.php on line 14
Anybody know why?
Thanks.
Ry
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[PHP-DB] forced page links...
I am wondering if there is a good tutorial on how to use php to make users come from a page before they can access another. For instance, say you have one page called index.php which contains a form that once filled in links to a confirmation page to verify the users data was entered correctly then from there links to a page that stores the users data into a database table. The problem I would like to alleviate is to make sure the users cannot just type in the name of the last page in the url and put a blank entry into the database. I have read a few tutorials on using sessions but I am unclear on how the use of sessions would require the user to visit each page in succession. Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: forced page links...
Can you accomplish this without having the user logged in? I am unconcerned
with whether or not the user has logged in, I simply need the user to not be
able to bypass a page with a form to go directly to the end page which will
connect and store the form contents into a db.
Thanks in advance,
Jas
Lerp [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi there :) I use session variables to do this, once the user has logged
in
and you know they are who they say they are, set a session variable using
session_register(isloggedin);
$isloggedin = yes;
And at the top of every page you want protected simply do a check to see
if
it is set or not. Below code checks to see if that var is set and if not
it
redirects them to another page.
?php session_start(); ?
?php
if (!isset($HTTP_SESSION_VARS[islogged])){
header(Location:index.php);
}
?
Hope this helps, Joe :)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
I am wondering if there is a good tutorial on how to use php to make
users
come from a page before they can access another. For instance, say you
have
one page called index.php which contains a form that once filled in
links
to
a confirmation page to verify the users data was entered correctly then
from
there links to a page that stores the users data into a database table.
The
problem I would like to alleviate is to make sure the users cannot just
type
in the name of the last page in the url and put a blank entry into the
database. I have read a few tutorials on using sessions but I am
unclear
on
how the use of sessions would require the user to visit each page in
succession. Thanks in advance,
Jas
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Re: [PHP-DB] Passing contents of array on as variables...
Would you have an example of that?
Jas
Beau Lebens [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
rather than all the hidden fields and stuff you guys are trying to do -
why
not just build the array variable you want, then save it into a session,
on
the next page, access the session variable and wallah - there's your
array,
still in tact and just as you left it (ie. containing all the info you
want,
not the word Array)
HTH
beua
// -Original Message-
// From: jas [mailto:[EMAIL PROTECTED]]
// Sent: Monday, 25 February 2002 5:50 PM
// To: [EMAIL PROTECTED]
// Subject: Re: [PHP-DB] Passing contents of array on as variables...
//
//
// Bjorn,
// I just wanted to thank you for giving me alot of insight
// into what I was
// trying to accomplish however, due to time restraints I have
// decided to do it
// a different way by pulling the entries into a select box
// that a customer can
// use to delete the different items. Once again, thanks a ton!!!!
// Jas
//
// Jas [EMAIL PROTECTED] wrote in message
// [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
// Ok, I think I understand what you are talking about. So
// now I know what I
// need to accomplish, and would you happen to know of a good
// place to learn
// about something like this? For some reason this function
// keeps seeming to
// elude me and if you know of a good tutorial on how to
// accomplish this I
// would appreciate it. Thanks again,
// Jas
// [EMAIL PROTECTED] wrote in message
// [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
// One problem I see is that you are sending all of the
// values for all of
// your
// fields (except the id field) from the first page to the
// second page, not
// just
// the ones that are checked, so even if it was working
// properly, you would
// get
// a list of all items in the table, not just the checked
// items. You need
// to
// send the array of id's (since this is the name for the
// checkbox) from
// the
// first page to the second one (again in the checkbox form
// element, you
// did
// not
// put a value on it).
//
// What your first page is currently doing is pulling all of the
// information
// out
// of the database. Then as soon as you pull each item out, you are
// putting
// it
// into an hidden form element array, ie. $car_type[],
// $car_model[], etc.
// regardless of whether the checkbox is checked or not to
// pass to the next
// page. You do not know if the checkbox is checked or not
// until the next
// page
// when it looks at the values in the id array. On the
// second page, you
// need
// to
// look at the id array and then (through a database call)
// pull the row
// from
// the
// table for each id in the array.
//
// HTH
//
// MB
//
// jas [EMAIL PROTECTED] said:
//
//$i=0;
//while
//
// ($car_type[$i],$car_model[$i],$car_year[$i],$car_price[$i],$c
// ar_vin[$i])
// {
//$i ++;
//}
//Is what I added and this is what is being output to
// the screen at this
//point...
//=0; while () { ++; }
//now i am still too new to php to understand why it is
// not putting the
//contents of the array into my hidden fields like it
// does on my first
// page (i
//can see them when I view source).
//[EMAIL PROTECTED] wrote in message
//[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
// When you call your $car_type in the second page, you
// need to set a
//variable
// starting at 0 and call it as $car_type[0] and loop
// through all of
// the
//values
// in the array.
//
// ie.
//
// $i=0;
// while ($car_type[$i]) {
//
// I have added more code below that should help.
//
// MB
//
//
// jas [EMAIL PROTECTED] said:
//
// Yeah, tried that and it still isnt passing the
// contents of the
// array
// as
//a
// varible to the confirmation page for deletion. I
// am at a loss on
// this
//one.
// [EMAIL PROTECTED] wrote in message
// [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
// You didn't add the value part of the hidden element, ie.
//
// INPUT TYPE=\hidden\ NAME=\car_type\
// value=\$myrow[car_type]\
//
// You were naming the field as an array with no value.
//
// if you want to pass this as an array of values,
// you would need
// to
// use:
//
// INPUT TYPE=\hidden\ NAME=\car_type[]\
// value=\$myrow[car_type]\
//
// Give that a try and see if it works.
//
// HTH
//
// MB
//
//
// jas [EMAIL PROTECTED] said:
//
//As of right now if you run the php script and
// view source on
// the
//page
// y
[PHP-DB] Passing contents of array on as variables...
Ok to this point I have been able to query a database table, display the results in tables and within a form. As of yet I have been able to insert the contents of the db query into an array for further processing. However, when I try to pass the contents of that array on to another page to delete selected records from the db it only displays the word array. Not quite sure where I need to go from here... Any insight would be a great help. Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Passing contents of array on as variables...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was getting confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
// Start to count number of records in selected table and loop until done
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$count ++;
// Begin placing them into an hidden fields and then array
echo
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
// Place items on separate cells in html table
trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
// Create checkbox so user can delete items if needed
echo /tdtdinput type=\checkbox\
name=\id[.$myrow[id].]\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
// End loop and print the infamous delete button
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Here is page two...
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm Record
Deletion/B/fonthr color=\33\/td
/tr
tr
td width=\30%\BType Of Car: /B/td
td$car_type/td // here is where it prints the word array instead of
type of car
/tr
tr
td width=\30%\BModel Of Car: /B/td
td$car_model/td
/tr
tr
td width=\30%\BYear Of Car: /B/td
td$car_year/td
/tr
tr
td width=\30%\BPrice Of Car: /B/td
td$car_price/td
/tr
tr
td width=\30%\BVIN Of Car: /B/td
td$car_vin/td
/tr
tr
td colspan=\3\hr color=\33\/td
/tr
tr
tdinput type=\submit\ name=\delete\ value=\delete\/td
/tr
/form
/table);
?
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Let's see what your code looks like now and where it is returning the
word array. That might help determine where the problem lies now.
MB
jas [EMAIL PROTECTED] said:
Ok to this point I have been able to query a database table, display the
results in tables and within a form. As of yet I have been able to
insert
the contents of the db query into an array for further processing.
However,
when I try to pass the contents of that array on to another page to
delete
selected records from the db it only displays the word array. Not
quite
sure where I need to go from here... Any insight would be a great help.
Thanks in advance,
Jas
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To unsubscribe, visit: http://www.php.net/unsub.php
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Passing contents of array on as variables...
As of right now if you run the php script and view source on the page you
can see that it does place the content of db table into array... ex.
BCurrent Inventory/B/fonthr color=33/td/tr
INPUT TYPE=hidden NAME=car_type[Ford]
INPUT TYPE=hidden NAME=car_model[Ranger]
INPUT TYPE=hidden NAME=car_year[1999]
INPUT TYPE=hidden NAME=car_price[5600]
INPUT TYPE=hidden NAME=car_vin[no vin]
trtd width=30%BType Of Car: /B/tdtdFord/tdtdinput
type=checkbox name=id[1]remove/td/tr
but on the following page (after selecting items to delete) it just displays
the word array for each field (i.e. car_type etc.)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was getting
confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
// Start to count number of records in selected table and loop until done
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$count ++;
// Begin placing them into an hidden fields and then array
echo
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
// Place items on separate cells in html table
trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
// Create checkbox so user can delete items if needed
echo /tdtdinput type=\checkbox\
name=\id[.$myrow[id].]\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
// End loop and print the infamous delete button
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Here is page two...
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm Record
Deletion/B/fonthr color=\33\/td
/tr
tr
td width=\30%\BType Of Car: /B/td
td$car_type/td // here is where it prints the word array instead of
type of car
/tr
tr
td width=\30%\BModel Of Car: /B/td
td$car_model/td
/tr
tr
td width=\30%\BYear Of Car: /B/td
td$car_year/td
/tr
tr
td width=\30%\BPrice Of Car: /B/td
td$car_price/td
/tr
tr
td width=\30%\BVIN Of Car: /B/td
td$car_vin/td
/tr
tr
td colspan=\3\hr color=\33\/td
/tr
tr
tdinput type=\submit\ name=\delete\ value=\delete\/td
/tr
/form
/table);
?
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Let's see what your code looks like now and where it is returning the
word array. That might help determine where the problem lies now.
MB
jas [EMAIL PROTECTED] said:
Ok to this point I have been able to query a database table, display
the
results in tables and within a form. As of yet I have been able to
insert
the contents of the db query into an array for further processing.
However,
when I try to pass the contents of that array on to another page to
delete
selected records from the db it only displays the word array. Not
quite
sure where I need to go from here... Any insight would be a great
help.
Thanks in advance,
Jas
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP-DB] Passing contents of array on as variables...
Yeah, tried that and it still isnt passing the contents of the array as a
varible to the confirmation page for deletion. I am at a loss on this one.
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
You didn't add the value part of the hidden element, ie.
INPUT TYPE=\hidden\ NAME=\car_type\ value=\$myrow[car_type]\
You were naming the field as an array with no value.
if you want to pass this as an array of values, you would need to use:
INPUT TYPE=\hidden\ NAME=\car_type[]\ value=\$myrow[car_type]\
Give that a try and see if it works.
HTH
MB
jas [EMAIL PROTECTED] said:
As of right now if you run the php script and view source on the page
you
can see that it does place the content of db table into array... ex.
BCurrent Inventory/B/fonthr color=33/td/tr
INPUT TYPE=hidden NAME=car_type[Ford]
INPUT TYPE=hidden NAME=car_model[Ranger]
INPUT TYPE=hidden NAME=car_year[1999]
INPUT TYPE=hidden NAME=car_price[5600]
INPUT TYPE=hidden NAME=car_vin[no vin]
trtd width=30%BType Of Car: /B/tdtdFord/tdtdinput
type=checkbox name=id[1]remove/td/tr
but on the following page (after selecting items to delete) it just
displays
the word array for each field (i.e. car_type etc.)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was getting
confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
// Start to count number of records in selected table and loop until
done
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$count ++;
// Begin placing them into an hidden fields and then array
echo
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
// Place items on separate cells in html table
trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
// Create checkbox so user can delete items if needed
echo /tdtdinput type=\checkbox\
name=\id[.$myrow[id].]\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr
color=\33\/td/tr\n;
}
// End loop and print the infamous delete button
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Here is page two...
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm
Record
Deletion/B/fonthr color=\33\/td
/tr
tr
td width=\30%\BType Of Car: /B/td
td$car_type/td // here is where it prints the word array
instead of
type of car
/tr
tr
td width=\30%\BModel Of Car: /B/td
td$car_model/td
/tr
tr
td width=\30%\BYear Of Car: /B/td
td$car_year/td
/tr
tr
td width=\30%\BPrice Of Car: /B/td
td$car_price/td
/tr
tr
td width=\30%\BVIN Of Car: /B/td
td$car_vin/td
/tr
tr
td colspan=\3\hr color=\33\/td
/tr
tr
tdinput type=\submit\ name=\delete\ value=\delete\/td
/tr
/form
/table);
?
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Let's see what your code looks like now and where it is returning
the
word array. That might help determine where the problem lies now.
MB
jas [EMAIL PROTECTED] said:
Ok to this point I have been able to query a database table,
display
th
Re: [PHP-DB] Passing contents of array on as variables...
$i=0;
while
($car_type[$i],$car_model[$i],$car_year[$i],$car_price[$i],$car_vin[$i]) {
$i ++;
}
Is what I added and this is what is being output to the screen at this
point...
=0; while () { ++; }
now i am still too new to php to understand why it is not putting the
contents of the array into my hidden fields like it does on my first page (i
can see them when I view source).
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
When you call your $car_type in the second page, you need to set a
variable
starting at 0 and call it as $car_type[0] and loop through all of the
values
in the array.
ie.
$i=0;
while ($car_type[$i]) {
I have added more code below that should help.
MB
jas [EMAIL PROTECTED] said:
Yeah, tried that and it still isnt passing the contents of the array as
a
varible to the confirmation page for deletion. I am at a loss on this
one.
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
You didn't add the value part of the hidden element, ie.
INPUT TYPE=\hidden\ NAME=\car_type\ value=\$myrow[car_type]\
You were naming the field as an array with no value.
if you want to pass this as an array of values, you would need to use:
INPUT TYPE=\hidden\ NAME=\car_type[]\ value=\$myrow[car_type]\
Give that a try and see if it works.
HTH
MB
jas [EMAIL PROTECTED] said:
As of right now if you run the php script and view source on the
page
you
can see that it does place the content of db table into array... ex.
BCurrent Inventory/B/fonthr color=33/td/tr
INPUT TYPE=hidden NAME=car_type[Ford]
INPUT TYPE=hidden NAME=car_model[Ranger]
INPUT TYPE=hidden NAME=car_year[1999]
INPUT TYPE=hidden NAME=car_price[5600]
INPUT TYPE=hidden NAME=car_vin[no vin]
trtd width=30%BType Of Car: /B/tdtdFord/tdtdinput
type=checkbox name=id[1]remove/td/tr
but on the following page (after selecting items to delete) it just
displays
the word array for each field (i.e. car_type etc.)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was getting
confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could
not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\
width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
// Start to count number of records in selected table and loop
until
done
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$count ++;
// Begin placing them into an hidden fields and then array
echo
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
// Place items on separate cells in html table
trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
// Create checkbox so user can delete items if needed
echo /tdtdinput type=\checkbox\
name=\id[.$myrow[id].]\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr
color=\33\/td/tr\n;
}
// End loop and print the infamous delete button
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Here is page two...
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\ NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\ NAME=\car_year[.$myrow[car_year].]\
INPUT TYPE=\hidden\ NAME=\car_price[.$myrow[car_price].]\
INPUT TYPE=\hidden\ NAME=\car_vin[.$myrow[car_vin].]\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm
Record
Deletion/B/fonthr color=\33\/td
/tr
Here is where
Re: [PHP-DB] Passing contents of array on as variables...
Ok, I think I understand what you are talking about. So now I know what I
need to accomplish, and would you happen to know of a good place to learn
about something like this? For some reason this function keeps seeming to
elude me and if you know of a good tutorial on how to accomplish this I
would appreciate it. Thanks again,
Jas
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
One problem I see is that you are sending all of the values for all of
your
fields (except the id field) from the first page to the second page, not
just
the ones that are checked, so even if it was working properly, you would
get
a list of all items in the table, not just the checked items. You need to
send the array of id's (since this is the name for the checkbox) from the
first page to the second one (again in the checkbox form element, you did
not
put a value on it).
What your first page is currently doing is pulling all of the information
out
of the database. Then as soon as you pull each item out, you are putting
it
into an hidden form element array, ie. $car_type[], $car_model[], etc.
regardless of whether the checkbox is checked or not to pass to the next
page. You do not know if the checkbox is checked or not until the next
page
when it looks at the values in the id array. On the second page, you need
to
look at the id array and then (through a database call) pull the row from
the
table for each id in the array.
HTH
MB
jas [EMAIL PROTECTED] said:
$i=0;
while
($car_type[$i],$car_model[$i],$car_year[$i],$car_price[$i],$car_vin[$i])
{
$i ++;
}
Is what I added and this is what is being output to the screen at this
point...
=0; while () { ++; }
now i am still too new to php to understand why it is not putting the
contents of the array into my hidden fields like it does on my first
page (i
can see them when I view source).
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
When you call your $car_type in the second page, you need to set a
variable
starting at 0 and call it as $car_type[0] and loop through all of the
values
in the array.
ie.
$i=0;
while ($car_type[$i]) {
I have added more code below that should help.
MB
jas [EMAIL PROTECTED] said:
Yeah, tried that and it still isnt passing the contents of the array
as
a
varible to the confirmation page for deletion. I am at a loss on
this
one.
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
You didn't add the value part of the hidden element, ie.
INPUT TYPE=\hidden\ NAME=\car_type\
value=\$myrow[car_type]\
You were naming the field as an array with no value.
if you want to pass this as an array of values, you would need to
use:
INPUT TYPE=\hidden\ NAME=\car_type[]\
value=\$myrow[car_type]\
Give that a try and see if it works.
HTH
MB
jas [EMAIL PROTECTED] said:
As of right now if you run the php script and view source on the
page
you
can see that it does place the content of db table into array...
ex.
BCurrent Inventory/B/fonthr color=33/td/tr
INPUT TYPE=hidden NAME=car_type[Ford]
INPUT TYPE=hidden NAME=car_model[Ranger]
INPUT TYPE=hidden NAME=car_year[1999]
INPUT TYPE=hidden NAME=car_price[5600]
INPUT TYPE=hidden NAME=car_vin[no vin]
trtd width=30%BType Of Car:
/B/tdtdFord/tdtdinput
type=checkbox name=id[1]remove/td/tr
but on the following page (after selecting items to delete) it
just
displays
the word array for each field (i.e. car_type etc.)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was
getting
confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or
die(Could
not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\
width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font
size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
// Start to count number of records in selected table and loop
until
done
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$count ++;
// Begin placing them into an hidden fields and then array
echo
INPUT TYPE=\hidden\
NAME=\car_type[.$myrow[car_type].]\
INPUT TYPE=\hidden\
NAME=\car_model[.$myrow[car_model].]\
INPUT TYPE=\hidden\
Re: [PHP-DB] Passing contents of array on as variables...
Bjorn,
I just wanted to thank you for giving me alot of insight into what I was
trying to accomplish however, due to time restraints I have decided to do it
a different way by pulling the entries into a select box that a customer can
use to delete the different items. Once again, thanks a ton
Jas
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok, I think I understand what you are talking about. So now I know what I
need to accomplish, and would you happen to know of a good place to learn
about something like this? For some reason this function keeps seeming to
elude me and if you know of a good tutorial on how to accomplish this I
would appreciate it. Thanks again,
Jas
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
One problem I see is that you are sending all of the values for all of
your
fields (except the id field) from the first page to the second page, not
just
the ones that are checked, so even if it was working properly, you would
get
a list of all items in the table, not just the checked items. You need
to
send the array of id's (since this is the name for the checkbox) from
the
first page to the second one (again in the checkbox form element, you
did
not
put a value on it).
What your first page is currently doing is pulling all of the
information
out
of the database. Then as soon as you pull each item out, you are
putting
it
into an hidden form element array, ie. $car_type[], $car_model[], etc.
regardless of whether the checkbox is checked or not to pass to the next
page. You do not know if the checkbox is checked or not until the next
page
when it looks at the values in the id array. On the second page, you
need
to
look at the id array and then (through a database call) pull the row
from
the
table for each id in the array.
HTH
MB
jas [EMAIL PROTECTED] said:
$i=0;
while
($car_type[$i],$car_model[$i],$car_year[$i],$car_price[$i],$car_vin[$i])
{
$i ++;
}
Is what I added and this is what is being output to the screen at this
point...
=0; while () { ++; }
now i am still too new to php to understand why it is not putting the
contents of the array into my hidden fields like it does on my first
page (i
can see them when I view source).
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
When you call your $car_type in the second page, you need to set a
variable
starting at 0 and call it as $car_type[0] and loop through all of
the
values
in the array.
ie.
$i=0;
while ($car_type[$i]) {
I have added more code below that should help.
MB
jas [EMAIL PROTECTED] said:
Yeah, tried that and it still isnt passing the contents of the
array
as
a
varible to the confirmation page for deletion. I am at a loss on
this
one.
[EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
You didn't add the value part of the hidden element, ie.
INPUT TYPE=\hidden\ NAME=\car_type\
value=\$myrow[car_type]\
You were naming the field as an array with no value.
if you want to pass this as an array of values, you would need
to
use:
INPUT TYPE=\hidden\ NAME=\car_type[]\
value=\$myrow[car_type]\
Give that a try and see if it works.
HTH
MB
jas [EMAIL PROTECTED] said:
As of right now if you run the php script and view source on
the
page
you
can see that it does place the content of db table into
array...
ex.
BCurrent Inventory/B/fonthr color=33/td/tr
INPUT TYPE=hidden NAME=car_type[Ford]
INPUT TYPE=hidden NAME=car_model[Ranger]
INPUT TYPE=hidden NAME=car_year[1999]
INPUT TYPE=hidden NAME=car_price[5600]
INPUT TYPE=hidden NAME=car_vin[no vin]
trtd width=30%BType Of Car:
/B/tdtdFord/tdtdinput
type=checkbox name=id[1]remove/td/tr
but on the following page (after selecting items to delete) it
just
displays
the word array for each field (i.e. car_type etc.)
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Oops... yeah its a monday.
Bjorn, I reverted back to my original code because I was
getting
confused...
Here is page one...
?php
// Database connection paramaters
require '../path/to/db.php';
// SQL statement to get current inventory
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or
die(Could
not
execute query, please try again later);
// Creating table to make data look pretty
echo table border=\0\ class=\table-body\
width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.p
Re: [PHP-DB] Passing contents of array on as variables...
Awesome... Thanks alot Natalie! I actually decided to pull the contents of the table into a select list as my option to delete the items, out of frustration. In any event I would love to take a look at your code to see a different approach to accomplishing my original problem. Thanks again everyone. Jas Natalie Leotta [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Jas, I've run into the same kind of problem with the site I'm doing for work. My first solution was to use the query string to pass contents of an array (array[0]array[1]...) but I got to the point where I was passing too much information and it wouldn't work. Now I use hidden data in my form and I dynamically name the fields in a loop. Then in the next page I use the get_post_vars and use substring so I can say if the first part of the var name is my array name then I get the position from the next part of the var name (array0, array1...) and use that position to put it in the new array. I definitely wouldn't assume it's the cleanest way to do it, but I didn't want to use sessions (if they'd even work) and the query string had some problems with confidential data anyway (the users can see it and we aren't supposed to display data in certain situations). If you'd like to go with this method and want some pointers let me know and I'll send you some of my code. Good luck! -Natalie -Original Message- From: [EMAIL PROTECTED] [SMTP:[EMAIL PROTECTED]] Sent: Monday, February 25, 2002 1:01 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Passing contents of array on as variables... Let's see what your code looks like now and where it is returning the word array. That might help determine where the problem lies now. MB jas [EMAIL PROTECTED] said: Ok to this point I have been able to query a database table, display the results in tables and within a form. As of yet I have been able to insert the contents of the db query into an array for further processing. However, when I try to pass the contents of that array on to another page to delete selected records from the db it only displays the word array. Not quite sure where I need to go from here... Any insight would be a great help. Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Associating table id...
Ok I have been working on this function to delete an item from a mysql
database and so far I have not had any success. Here is the problem I am
having (after alot of headaches), I need to be able to associate a checkbox
to a record or a set of records... For example, page 1 queries a database
and pulls the results into a table and in that table is a form with a
checkbox for each record pulled from said database... code is here...
?php
require '../path/to/db.php'; //connection script
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\id\
value=\id\remove/td/tr\n;
echo rest of fields in database;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Now what I need to do is to have the checkbox item be associated with all
one record set. Table structure is here
-
| id | car_model | car_type | car_year | car_price | car_vin | dlr_num |
-
| 0 | ford | bronco| 1997 | 6700 | vin# | dlr
# |
-
etc. etc.
From the first page it links to page which queries the db and deletes the
selected records... however, I am not able to get the id to be associated
with the records in the table... the deletion script is as follows
?php
require '../path/to/db.php';
$table_name = cur_inv;
$sql = DELETE FROM $table_name WHERE id = '$id';
echo($sql);
$result = mysql_query($sql,$dbh) or die(mysql_error());
print(record deleted);
?
If anyone has ever run into this please help... I am still fairly new to php
and mysql and there is definately something I am missing here and I think
its because I need to associate the checkbox with the id and the id field
from the database does not seem to be linked to the rest of the record in
said fields...
Thanks in advance,
Jas
And yes I do know I posted the same question yesterday... I still cannot
figure this one out.
--
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[PHP-DB] Re: Associating table id...
I sincerely wish someone could give me some insight as to what I am doing
wrong here. I have been scouring php.net for any website that has a good
tutorial or article on how to pass the variables from page to page while
assigning a variable a whole record set from a database. If anyone has run
into the same problem that I am having please give me a shove in the right
direction. Thanks in advance,
Jas
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok I have been working on this function to delete an item from a mysql
database and so far I have not had any success. Here is the problem I am
having (after alot of headaches), I need to be able to associate a
checkbox
to a record or a set of records... For example, page 1 queries a database
and pulls the results into a table and in that table is a form with a
checkbox for each record pulled from said database... code is here...
?php
require '../path/to/db.php'; //connection script
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\id\
value=\id\remove/td/tr\n;
echo rest of fields in database;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Now what I need to do is to have the checkbox item be associated with all
one record set. Table structure is here
-
| id | car_model | car_type | car_year | car_price | car_vin | dlr_num |
-
| 0 | ford | bronco| 1997 | 6700 | vin# |
dlr
# |
-
etc. etc.
From the first page it links to page which queries the db and deletes the
selected records... however, I am not able to get the id to be
associated
with the records in the table... the deletion script is as follows
?php
require '../path/to/db.php';
$table_name = cur_inv;
$sql = DELETE FROM $table_name WHERE id = '$id';
echo($sql);
$result = mysql_query($sql,$dbh) or die(mysql_error());
print(record deleted);
?
If anyone has ever run into this please help... I am still fairly new to
php
and mysql and there is definately something I am missing here and I think
its because I need to associate the checkbox with the id and the id field
from the database does not seem to be linked to the rest of the record in
said fields...
Thanks in advance,
Jas
And yes I do know I posted the same question yesterday... I still cannot
figure this one out.
--
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[PHP-DB] Array HELL!!!!
I don't know what it is but I am having a hell of a time trying to get some
results of a query setup into an array or variable (too much of a newbie to
know which) that can be passed to a confirmation page before deleting the
record from a table. I have given up working on this but for those of you
that want to finish it here is the code and the table structure...
[Table Structure]
id int(30) DEFAULT '0' NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id)
[Page 1 - Queries DB table for records]
?php
require '../path/to/db.php';
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$id = $row[id];
$car_type = $row[car_type];
$car_model = $row[car_model];
$car_year = $row[car_year];
$car_price = $row[car_price];
$car_vin = $row[car_vin];
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\id[]\
value=\.$myrow[id].\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
[Page 2 - Takes records and confirms which ones to be deleted]
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\id\ VALUE=\$id\
INPUT TYPE=\hidden\ NAME=\car_type\ VALUE=\$car_type\
INPUT TYPE=\hidden\ NAME=\car_model\ VALUE=\$car_model\
INPUT TYPE=\hidden\ NAME=\car_year\ VALUE=\$car_year\
INPUT TYPE=\hidden\ NAME=\car_price\ VALUE=\$car_price\
INPUT TYPE=\hidden\ NAME=\car_vin\ VALUE=\$car_vin\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm Record
Deletion/B/fonthr color=\33\/td
/tr
tr
td width=\30%\BType Of Car: /B/td
td$car_type/td
/tr
tr
td width=\30%\BModel Of Car: /B/td
td$car_model/td
/tr
tr
td width=\30%\BYear Of Car: /B/td
td$car_year/td
/tr
tr
td width=\30%\BPrice Of Car: /B/td
td$car_price/td
/tr
tr
td width=\30%\BVIN Of Car: /B/td
td$car_vin/td
/tr
tr
td colspan=\3\hr color=\33\/td
/tr
tr
tdinput type=\submit\ name=\delete\ value=\delete\/td
/tr
/form
/table);
?
[Page 3 - Connects to DB and deletes selected records]
?php
require '../path/to/db.php';
$table_name = cur_inv;
$sql = DELETE FROM $table_name WHERE id = '$id';
echo($sql);
$result = mysql_query($sql,$dbh) or die(mysql_error());
print(body bgcolor=\ff9900\p class=\done\You have successfully
removed your items from the database.);
?
If anyone has the time to finish it or give me some pointers on what the
hell I am doing wrong please let me know, I would be very glad to hear your
opinion and the correct way to accomplish this. (And, yes I went through
just about every tutorial I could find on how to delete records, which by
the way did nothing for putting the results of a select statement into an
array or variable for futher processing.) Have a good weekend everyone!!!
Pissed and frustrated...
Jas
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Re: [PHP-DB] Array HELL!!!!
Well ok now that I know it isn't an array I am having problems with, how
exactly am I supposed to create a variable (that can be passed to another
page) from a selected set of records?
Thanks again for the politeness,
Jas
Spyproductions Support Team [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
That's not Array Hell, this is Array Hell:
$Hell = array(satan,fire,brimstone);
-Mike
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 22, 2002 4:37 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Array HELL
I don't know what it is but I am having a hell of a time trying
to get some
results of a query setup into an array or variable (too much of a
newbie to
know which) that can be passed to a confirmation page before deleting
the
record from a table. I have given up working on this but for those of
you
that want to finish it here is the code and the table structure...
[Table Structure]
id int(30) DEFAULT '0' NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id)
[Page 1 - Queries DB table for records]
?php
require '../path/to/db.php';
$result = @mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\rem_conf.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_array($result)) {
$id = $row[id];
$car_type = $row[car_type];
$car_model = $row[car_model];
$car_year = $row[car_year];
$car_price = $row[car_price];
$car_vin = $row[car_vin];
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\id[]\
value=\.$myrow[id].\remove/td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr
color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
[Page 2 - Takes records and confirms which ones to be deleted]
?php
print(
table border=\0\ class=\table-body\ width=\100%\
form name=\rem_inv\ method=\post\ action=\done2.php3\
INPUT TYPE=\hidden\ NAME=\id\ VALUE=\$id\
INPUT TYPE=\hidden\ NAME=\car_type\ VALUE=\$car_type\
INPUT TYPE=\hidden\ NAME=\car_model\ VALUE=\$car_model\
INPUT TYPE=\hidden\ NAME=\car_year\ VALUE=\$car_year\
INPUT TYPE=\hidden\ NAME=\car_price\ VALUE=\$car_price\
INPUT TYPE=\hidden\ NAME=\car_vin\ VALUE=\$car_vin\
tr
td align=\center\ colspan=\3\font size=\4\BConfirm
Record
Deletion/B/fonthr color=\33\/td
/tr
tr
td width=\30%\BType Of Car: /B/td
td$car_type/td
/tr
tr
td width=\30%\BModel Of Car: /B/td
td$car_model/td
/tr
tr
td width=\30%\BYear Of Car: /B/td
td$car_year/td
/tr
tr
td width=\30%\BPrice Of Car: /B/td
td$car_price/td
/tr
tr
td width=\30%\BVIN Of Car: /B/td
td$car_vin/td
/tr
tr
td colspan=\3\hr color=\33\/td
/tr
tr
tdinput type=\submit\ name=\delete\ value=\delete\/td
/tr
/form
/table);
?
[Page 3 - Connects to DB and deletes selected records]
?php
require '../path/to/db.php';
$table_name = cur_inv;
$sql = DELETE FROM $table_name WHERE id = '$id';
echo($sql);
$result = mysql_query($sql,$dbh) or die(mysql_error());
print(body bgcolor=\ff9900\p class=\done\You have successfully
removed your items from the database.);
?
If anyone has the time to finish it or give me some pointers on what the
hell I am doing wrong please let me know, I would be very glad to
hear your
opinion and the correct way to accomplish this. (And, yes I went
through
just about every tutorial I could find on how to delete records, which
by
the way did nothing for putting the results of a select statement into
an
array or variable for futher processing.) Have a good weekend
everyone!!!
Pissed and frustrated...
Jas
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[PHP-DB] Parse Error
I hate to post this again but I have looked in a couple of php and mysql books but cannot seem to figure this one out. I am getting a parse error when trying to use php to delete records from a table. The error I am recieving is as follows Parse error: parse error in /path/to/php/done2.php3 on line 22 Here is the file that is giving me the error, any help would be great... I think that my problem is that I left out a variable to hold the $cars data but I am not sure. Here is the code... $db_name = test; $table_name = inventory; $connection = @mysql_connect(localhost, root, password) or die (Could not connect to database. Please try again later.); $db = @mysql_select_db($db_name,$connection) or die (Could not select database table. Please try again later.); $sql = DELETE FROM $table_name WHERE $cars = \$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl r_num\); $result = @mysql_query($sql, $connection) or die (Could not execute query. Any insight would be great... thanks again. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
After this file is pulled and the user selects which record to delete it
jumps to the other snippit of code in done2.php3 which is
$db_name = test;
$table_name = inventory;
$connection = @mysql_connect(localhost, root, password) or die
(Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = DELETE FROM $table_name WHERE $cars =
\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl
r_num\);
$result = @mysql_query($sql, $connection) or die (Could not execute
query.
Any insight would be great... thanks again.
Jas
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Re: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id)
test VALUES ('1', 'Ford', 'Bronco', '1969', '4500', 'vin897655', 'none');
Sorry about that... still a bit of a newbie here. =)
Jas
Gurhan Ozen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Jas..
First of all you don't have the variable $cars assigned..
Second of all, we still don't know what the test table looks like? Does
it
only have one column with all the infos about cars populated in it???
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 12:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
After this file is pulled and the user selects which record to delete it
jumps to the other snippit of code in done2.php3 which is
$db_name = test;
$table_name = inventory;
$connection = @mysql_connect(localhost, root, password) or die
(Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not
select
database table. Please try again later.);
$sql = DELETE FROM $table_name WHERE $cars =
\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl
r_num\);
$result = @mysql_query($sql, $connection) or die (Could not execute
query.
Any insight would be great... thanks again.
Jas
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Re: [PHP-DB] Parse Error
Could you maybe give me an example of how to associate the checkbox with the
id? Thanks again,
Jas
Gurhan Ozen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok then it sure will not work.. because you have :
DELETE FROM $table_name WHERE $cars
=\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$d
lr_num\);
1- $cars is set to the value checkbox which doesn't exist in your test
table
2- The end of the statement $dlr_num\); is not right , that
parenthesis isn't supposed to be there
3- The delete statement is wrong...You are storing all that info in the
separate fields but you are giving all combined as a condition in only one
field ($cars).
3- Rewrite your delete statement to delete the rows identified by the id
column as DELETE FROM $table WHERE id ='id'; Of course in the page where
you list the cars to be deleted associate each checkbox with the car's id.
Hope this helps..
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 1:02 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id)
test VALUES ('1', 'Ford', 'Bronco', '1969', '4500', 'vin897655', 'none');
Sorry about that... still a bit of a newbie here. =)
Jas
Gurhan Ozen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Jas..
First of all you don't have the variable $cars assigned..
Second of all, we still don't know what the test table looks like?
Does
it
only have one column with all the infos about cars populated in it???
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 12:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to
delete
the items in the db. I dont know if this will help but like I said
before
any insight would be great. Thanks in advance.
Jas
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr
color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
After this file is pulled and the user selects which record to delete it
jumps to the other snippit of code in done2.php3 which is
$db_name = test;
$table_name = inventory;
$connection = @mysql_connect(localhost, root, password) or die
(Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not
select
database table. Please try again later.);
$sql = DELETE FROM $table_name WHERE $cars =
\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl
r_num\);
$result = @mysql_query($sql, $connection) or die (Could not execute
query.
Any insight would be great... thanks again.
Jas
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[PHP-DB] debugging?
Can someone tell me how I can find out why I am getting errors executing
queries when I try to delete items from a table? I have 2 files...
file 1. - Queries database, displays results with option to delete record
using a check box, code is as follows...
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\.$myrow[id].\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
file 2. - takes variable of checkbox and then queries db trying to delete
any record based on the id. code is as follows...
?php
require '../scripts/db.php';
$sql = @mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or
die(Could not execute query, please try again later);
print(changes made);
?
I have been trying everything under the sun and cannot get this to work.
please help.
Jas
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[PHP-DB] Errors Deleting...
Can someone tell me how I can find out why I am getting errors executing
queries when I try to delete items from a table? I have 2 files...
file 1. - Queries database, displays results with option to delete record
using a check box, code is as follows...
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\.$myrow[id].\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
file 2. - takes variable of checkbox and then queries db trying to delete
any record based on the id. code is as follows...
?php
require '../scripts/db.php';
$table_name = cur_inv;
$sql = @mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or
die(Could not execute query, please try again later);
print(changes made);
?
I have been trying everything under the sun and cannot get this to work.
please help.
Jas
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Re: [PHP-DB] Errors Deleting...
I wish I could get this to work... it has not been able to work at all. I
made the changes you have suggested and tried different variations
concerning the results items and so far no dice.
Here is my delete sql statement as of right now...
?php
require '../scripts/db.php';
$table_name = cur_inv;
$sql = DELETE FROM $table_name WHERE $id = 'id';
$result = @mysql_query($sql, $dbh) or die(Could not execute query, please
try again later);
?
I keep getting an error on the result portion, I have double, and triple
checked my table names, my db connection (which has error checking on it so
I know I am connecting fine). I am wondering would it be beneficial to
actually put a select statement in for the database table? I would think
declaring the variable would work fine. In any event I also changed the
checkbox name to id to match the variable in the delete statement. For
readability sake here is the form code as well
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\id\
value=\.$myrow[id].\remove/td
/tr\n;
echo rest of table and cell output;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Gurhan Ozen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Jas,
You still have cars as your checkbox name.. change it to id..
and make sure that you assign $table_name to the table name you want to
delete from in your done2.php3 file..
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 3:28 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Errors Deleting...
Can someone tell me how I can find out why I am getting errors executing
queries when I try to delete items from a table? I have 2 files...
file 1. - Queries database, displays results with option to delete record
using a check box, code is as follows...
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\.$myrow[id].\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
file 2. - takes variable of checkbox and then queries db trying to delete
any record based on the id. code is as follows...
?php
require '../scripts/db.php';
$table_name = cur_inv;
$sql = @mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or
die(Could not execute query, please try again later);
print(changes made);
?
I have been trying everything under the sun and cannot get this to work.
please help.
Jas
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Re: [PHP-DB] debugging?
Ok now that I can see the error message how can I fix it... the syntax looks correct and the error I am recieving is as follows You have an error in your SQL syntax near '= 'id'' at line 1 And this is my statement, $sql = mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or die(mysql_error()); Is there anywhere on php.net or mysql.org that give exact error meanings? Thanks again, Jas Olinux [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... jas, Try this: - Remove the @, it suppresses the error - change die(Could not execute query, please try again later); to die(mysql_error()); $sql = mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or die(mysql_error()); olinux --- jas [EMAIL PROTECTED] wrote: Can someone tell me how I can find out why I am getting errors executing queries when I try to delete items from a table? I have 2 files... file 1. - Queries database, displays results with option to delete record using a check box, code is as follows... ?php snip __ Do You Yahoo!? Yahoo! Sports - Coverage of the 2002 Olympic Games http://sports.yahoo.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] debugging?
Ok now that the sql statement is working I am getting an error on my result function... here is the result function $result = mysql_query($sql, $dbh) or die(mysql_error()); and here is the error I am recieving You have an error in your SQL syntax near '1, 1' at line 1 I dont know enough about php and mysql to understand what this means... is there some kind of chart or definitions for error messages? I tried looking on mysql.org and couldnt find anything specific. In any event, thanks in advance. Jas [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Here is what your delete statement should look like: $sql = mysql_query(DELETE FROM $table_name WHERE id = '$id',$dbh) or die(mysql_error()); Note the field name without the $ in front of it and the variable you are comparing it to with the $. One other note, in your table, you have id as a varchar(30) auto_increment. As far as I know, auto_increment can only be used on INT() field types, unless I missed something there. If you do end up changing the field type to INT, make sure you take the '' around the $id out. HTH MB jas [EMAIL PROTECTED] said: Ok now that I can see the error message how can I fix it... the syntax looks correct and the error I am recieving is as follows You have an error in your SQL syntax near '= 'id'' at line 1 And this is my statement, $sql = mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or die(mysql_error()); Is there anywhere on php.net or mysql.org that give exact error meanings? Thanks again, Jas Olinux [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... jas, Try this: - Remove the @, it suppresses the error - change die(Could not execute query, please try again later); to die(mysql_error()); $sql = mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or die(mysql_error()); olinux --- jas [EMAIL PROTECTED] wrote: Can someone tell me how I can find out why I am getting errors executing queries when I try to delete items from a table? I have 2 files... file 1. - Queries database, displays results with option to delete record using a check box, code is as follows... ?php snip __ Do You Yahoo!? Yahoo! Sports - Coverage of the 2002 Olympic Games http://sports.yahoo.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Deleting records...
Ok I am trying to delete records from a database using an html form. I have
2 files, file one = remove.php3 which queries the database table and
displays the current contents. Code is as follows.
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Once it queries the database it gives the user an option to check the
checkbox and delete that record. Here is the done2.php3 page that actually
contains the delete sql statement
$db_name = test;
$table_name = inventory;
$connection = @mysql_connect(localhost, root, password) or die (Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = DELETE FROM $table_name WHERE $cars =
\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl
r_num\);
$result = @mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
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[PHP-DB] Re: Deleting records...
I forgot to include the error message I am recieving when trying to remove
records using the check box.
Parse error: parse error in /path/to/php/done2.php3 on line 22
Thanks in advance.
Jas
Jas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok I am trying to delete records from a database using an html form. I
have
2 files, file one = remove.php3 which queries the database table and
displays the current contents. Code is as follows.
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\3\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\cars\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
?
Once it queries the database it gives the user an option to check the
checkbox and delete that record. Here is the done2.php3 page that
actually
contains the delete sql statement
$db_name = test;
$table_name = inventory;
$connection = @mysql_connect(localhost, root, password) or die
(Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = DELETE FROM $table_name WHERE $cars =
\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dl
r_num\);
$result = @mysql_query($sql, $connection) or die (Could not execute
query.
Please try again later.);
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[PHP-DB] formating problem
I feel kinda dumb for posting this but here it is... I am trying to query a
table then format the results so it looks like the rest of the site and the
darn delete button to remove db entries keeps showing up at the top. Here
is the code... Thanks in advance.
Jas
?php
require '../scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\table-body\ width=\100%\form
name=\rem_inv\ method=\post\ action=\done2.php3\
trtd align=\center\ colspan=\2\font size=\4\BCurrent
Inventory/B/fonthr color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /tdtdinput type=\checkbox\ name=\car_type\
value=\checkbox\remove/td
/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\3\hr color=\33\/td/tr\n;
}
echo input type=\submit\ name=\delete\
value=\delete\/form/table;
?
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[PHP-DB] optimizing script...
I have a script that prevents invalid characters from being passed from a
html form to a mysql database using php and I am wondering if there is a way
to condense it so instead of declaring the ereg_replace for each field in
the html form, to simply use one function to remove invalid characters and
html markup from all the fields in the form at once... here is my code, any
insight would be great. I am not asking you to do it for me, I just want to
know how and what the best way to condense it would be. Thanks in advance.
Jas
# - Error correction for car_type field
$car_type = ereg_replace (\, q, $car_type);
$car_type = ereg_replace (', a, $car_type);
$car_type = ereg_replace (%, p, $car_type);
$car_type = ereg_replace (, bs, $car_type);
# - Error correction for car_model field
$car_model = ereg_replace (\, q, $car_model);
$car_model = ereg_replace (', a, $car_model);
$car_model = ereg_replace (%, p, $car_model);
$car_model = ereg_replace (, bs, $car_model);
# - Error correction for car_year field
$car_year = ereg_replace (\, q, $car_year);
$car_year = ereg_replace (', a, $car_year);
$car_year = ereg_replace (%, p, $car_year);
$car_year = ereg_replace (, bs, $car_year);
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[PHP-DB] formating w/ table
I am having a little problem formating data retrieved from a database into
table cells... So far after connecting and querying the database table the
data is put into table cells, however after the first entry is displayed in
the table all the other entries loose the formating, HELP?
?php
require 'scripts/db.php';
$result = mysql_query(SELECT * FROM cur_inv,$dbh) or die(Could not
execute query, please try again later);
echo table border=\0\ class=\newhead\ width=\100%\trtd
align=\center\ colspan=\2\BCurrent Inventory/Bhr
color=\33\/td/tr;
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo trtd width=\30%\BType Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_type));
echo /td/tr\n;
echo trtd width=\30%\BModel Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_model));
echo /td/tr\n;
echo trtd width=\30%\BYear Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_year));
echo /td/tr\n;
echo trtd width=\30%\BPrice Of Car: /B/tdtd$;
printf(mysql_result($result,$count,car_price));
echo /td/tr\n;
echo trtd width=\30%\BVIN Of Car: /B/tdtd;
printf(mysql_result($result,$count,car_vin));
echo /td/trtrtd colspan=\2\hr
color=\33\/td/tr/table\n;
}
?
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[PHP-DB] Disable Right click w/ php?
I have been looking on php.net for a function to disallow people to right click to view source... anyone have a good idea of how to accomplish this without using java-script? Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Form Validation
Anyone know of a good function to strip characters and stuff that would cause mysql to crash from forms? Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Required pages...
I am wondering if there is a way to force users to come from a certain page. For an example I am using a login page which once authenticated allows users to change the contents of a web site without knowing alot of code etc. What I would like to do is make sure that the content management system will not be accessed unless the user logs in. I am certain sessions is the way to go on this, however I am still new enough to not understand exactly how they work and how to impliment them on a site. I have read a little bit on a tutorial on php.net. If anyone can give me an example of how this could be accomplished I would appriciate it. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Updating Database problem...
I am having a problem with a script that simply updates a few fields in a
database however, I am having a problem having the script to not update the
database fields if the form is invalid... Here is my script, I don't want
you to fix it for me unless you show me where I am going wrong or can point
me to a good tutorial on this type of function. Thanks in advance,
Jas
?php
# trim extra spaces from these variables
$c_name = trim($c_name);
$s_addy = trim($s_addy);
$city = trim($city);
$state = trim($state);
$zip = trim($zip);
$phone = trim($phone);
if($c_name == )
{
$c_nameerr = Please enter your Company Namebr;
$formerr = $formerr + 1;
}
if($s_addy == )
{
$s_addyerr = Please enter your Street Addressbr;
$formerr = $formerr + 1;
}
if($city == )
{
$cityerr = Please enter your Citybr;
$formerr = $formerr + 1;
}
if($state == )
{
$stateerr = Please enter your Statebr;
$formerr = $formerr + 1;
}
if($zip == )
{
$ziperr = Please enter your Zip Codebr;
$formerr = $formerr + 1;
}
if($phone == )
{
$phoneerr = Please enter your Phone Numberbr;
$formerr = $formerr + 1;
}
# - connects to the mysql database to edit information
$db_name = test;
$table_name = www_demo;
$connection = @mysql_connect(localhost, user, password) or die (Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET
c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\,zip=\z
ip\,phone=\$phone\;
$result = @mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
# --
?
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Re: [PHP-DB] Updating Database problem...
Sorry, but I figured it out. You were right, I do need to call the sql
update function after the error field returns a 0. Here is the code if any
one else out there is having the same problem.
?php
# trim extra spaces from these variables
$c_name = trim($c_name);
$s_addy = trim($s_addy);
$city = trim($city);
$state = trim($state);
$zip = trim($zip);
$phone = trim($phone);
if($c_name == )
{
$c_nameerr = Please enter your Company Namebr;
$formerr = $formerr + 1;
}
if($s_addy == )
{
$s_addyerr = Please enter your Street Addressbr;
$formerr = $formerr + 1;
}
if($city == )
{
$cityerr = Please enter your Citybr;
$formerr = $formerr + 1;
}
if($state == )
{
$stateerr = Please enter your Statebr;
$formerr = $formerr + 1;
}
if($zip == )
{
$ziperr = Please enter your Zip Codebr;
$formerr = $formerr + 1;
}
if($phone == )
{
$phoneerr = Please enter your Phone Numberbr;
$formerr = $formerr + 1;
}
?
- Then in body of document --
?php
if($formerr 0)
{
print('You need to fill out the form completely. Click the back button on
your browser to make the necessary changes.br');
echo $c_nameerr;
echo $s_addyerr;
echo $cityerr;
echo $stateerr;
echo $ziperr;
echo $phoneerr;
}
if($formerr == 0)
{
print(pYour changes have been saved to the database/p);
# - connects to the mysql database to edit information
$db_name = test;
$table_name = www_demo;
$connection = @mysql_connect(localhost, user, password) or die (Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET
c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\,zip=\z
ip\,phone=\$phone\;
$result = @mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
# --
}
?
Rick Emery [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Jas,
Why are you initiating an UPDATE command when you've not changed the
information you wish to update?
For instance, if $c_name is blank, you display a message to enter the
data;
yet, you DO NOT ASK the user to enter the correct information. Then, you
do
an UPDATE statment to enter that blank info into the record.
You also update $formerr, yet do not use it.
I've got a feeling you're not sharing all your code. Please do so if we
are
to help.
You also need a WHERE clause in you UPDATE; otherwise, ALL records will be
changed
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 31, 2002 11:14 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Updating Database problem...
I am having a problem with a script that simply updates a few fields in a
database however, I am having a problem having the script to not update
the
database fields if the form is invalid... Here is my script, I don't want
you to fix it for me unless you show me where I am going wrong or can
point
me to a good tutorial on this type of function. Thanks in advance,
Jas
?php
# trim extra spaces from these variables
$c_name = trim($c_name);
$s_addy = trim($s_addy);
$city = trim($city);
$state = trim($state);
$zip = trim($zip);
$phone = trim($phone);
if($c_name == )
{
$c_nameerr = Please enter your Company Namebr;
$formerr = $formerr + 1;
}
if($s_addy == )
{
$s_addyerr = Please enter your Street Addressbr;
$formerr = $formerr + 1;
}
if($city == )
{
$cityerr = Please enter your Citybr;
$formerr = $formerr + 1;
}
if($state == )
{
$stateerr = Please enter your Statebr;
$formerr = $formerr + 1;
}
if($zip == )
{
$ziperr = Please enter your Zip Codebr;
$formerr = $formerr + 1;
}
if($phone == )
{
$phoneerr = Please enter your Phone Numberbr;
$formerr = $formerr + 1;
}
# - connects to the mysql database to edit information
$db_name = test;
$table_name = www_demo;
$connection = @mysql_connect(localhost, user, password) or die
(Could
not connect to database. Please try again later.);
$db = @mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET
c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\,zip=\z
ip\,phone=\$phone\;
$result = @mysql_query($sql, $connection) or die (Could not execute
query.
Please try again later.);
# --
?
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[PHP-DB] query error...
Ok what is wrong with this sql statement? $sql = UPDATE $table_name SET c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\zip=\zi p\,phone=\$phone\; The error I recieve is as follows Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /path/to/wwwdemo_change.php3 on line 22 Parse error: parse error in /path/to/wwwdemo_change.php3 on line 22 I have looked on MySQL.com at the update function and it states I can use the SET under UPDATE for multiple fields separated by a , and it is not working... Any insight would be great. Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Combined sql statement
I would like to know how to have a php script loop through two different sql statements and use one according to a yes or no answer. I am still kinda new to php and I already have my sql statements, but I dont know how to use php to tell it to use one of the other. My sql statements are as follows... $sql = INSERT INTO $table_name (our_serv) VALUES (\$our_serv\) ; $sql = UPDATE $table_name SET our_serv=\$our_serv\; How can I get it to pick one of these... for instance if there is no entry in the unique table to enter one using the first sql statement, or if there is already an entry to simply update and overwrite the current table entry. Any help would be appriciated and if you could please document it so I understand and dont have to ask again. Thanks, jas WOW, your neat -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] phpMyAdmin Problem....
yeah that sounds like apache not being setup right Matt Williams [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I just added phpMyAdmin to a new Apache Server, and I'm getting this error: cannot load MySQL extension, please check PHP Configuration. I did some research on the web, but couldn't come up with a solution to this error and couldn't find anyone listing it. Does anyone know if this is a problem with conf.inc.php in phpMyAdmin or with the MySQL set up in PHP on the server? Any suggestions from there? Sound like you need to recompile with mysql support. check phpinfo() to see if mysql is missing. m: -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] DB Connection Class
Ok, I hate to ask this again but I am having a hard time developing a connection class to be reused. I need to be able to just connect to the database for several pages because I have multiple queries per page. Unfortunately I do not know enough about php to effictively do this. I have read a few tutorials on the subject but they keep going way over my head, I need a simple to understand method of reusing a connection for each query. Please help! Thanks in advance! Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] DB Connections
Ok, I am having a hard time coming up with a mysql connection class that simply allows me to setup a require(db_connection.inc) at the top of my php pages that allows a consistent connection to the mysql db and allows multiple queries thoughout the php page in question. I have read a few tutorials on it and as of yet they all seem to go over my head. If someone could give me some more insight into this area I would greatly appriciate it. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] MySQL Connections
Ok how can I find some good resources on how to create a db connection class that doesnt go over my head? I need to be able to connect to a mysql database as the page loads and thoughout the page run 6 separate queries on different fields. please help! jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Connection class
Well it was acutally so that I can create a content management system for my employer. Another gentleman helped me with it and it is working great I simply created a file *.php with all my connection commands then did a require on the page(s) that needed to query the db to pull information about what is currently inserted into different fields. My NEW problem has to do with getting my INSERT statements to overwrite the distinctive fields when storing updates etc. I am looking into how to accomplish this at the moment. Sorry again for posting the same question a few times, it seems my email editor choked. =) Thanks again. Jas Rick Emery [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... huh? You open the connection once for the page and then you CAN make multiple queries to it from the same page. Why are you trying to create a class to do this (unless it's simply an intellectual exercise to create a class)? -Original Message- From: jas [mailto:[EMAIL PROTECTED]] Sent: Saturday, January 26, 2002 12:08 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Connection class Ok well I am in a little over my head with a connection class that I can use over and over for multiple queries against a mysql db on one page. I have read a few tutorials on the subject of how to use class files but I still dont understand php enough to know how to make a connection class that simply allows me to connect to a mysql database once the page loads and keeps the connection to run multiple queries on the same page. HELP!!! Thanks again in advance... Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] MySQL Connection Class
I know this may seem a little vague but I would like to know where a good tutorial on creating database connection class files would be. I have been looking and as of yet I have not found one that deals specifically with this question. Thanks in advance. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
