Re: [PHP] mysql problems [SOLVED]
Sean Greenslade wrote:
[MASSIVE
SNIP]
Well, from what I saw while wading through your
code, you allow
unsanitized
variables to be
concatenated to your queries. Big no-no! For ANY
client-generated variable, always sanitize with
mysql_real_escape_string.
In
fact, sanitize all your
variables. It can't hurt.
Also, please don't take a
request for your entire code too literally. We
don't like to see
pages and pages and pages of code, just the pertinent
bits.
--
--Zootboy
Sent from my PC.
Thanks to all, but it was an infinite loop. there was a
while ($_parent != 0) { } loop. In the loop the database
is queried. If the returned number of rows is greater than 0 then
perform then grab a $_parent from the database. At some point, there
must be a parent that is = 0 and the loop breaks. However, if the
page is called with category number that doesn't exist, then the if/then
clause is never true and $_parent never gets set to 0. I simply
added and else clause.
while ($_parent != 0)
{
if
($num_rows 0)
{
perform some action
}
else
{
$_parent =
0;
}
}
and that solved the
problem.
Thank you, everyone for your help.
Curtis
RE: [PHP] mysql problems [SOLVED]
[SNIP]
added and else clause.
while ($_parent != 0)
{
if
($num_rows 0)
{
perform some action
}
else
{
$_parent =
0;
}
}
and that solved the
problem.
Thank you, everyone for your help.
Curtis
A small remark:
I think it is good programming practice to place such static if-clauses before
the while statement.
This prevents a lot of redundant checks and thus saves time.
Best regards,
Jasper
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Re: Re: [PHP] mysql problems
On 11 May 2011 at 19:25, Curtis Maurand cur...@maurand.com wrote: $_cartTotal=$0.00; Surely that should be: $_cartTotal = 0.00; tim -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: Re: [PHP] mysql problems
Tim Streater wrote: On 11 May 2011 at 19:25, Curtis Maurand cur...@maurand.com wrote: $_cartTotal=$0.00; Surely that should be: $_cartTotal = 0.00; Good pickup. I missed that. I didn't write the code, I'm just trying to figure out what's going on. Thanks, I'll look at that. --C
[PHP] mysql problems
I'm running PHP 5.3.6, Mysql 5.1.51 and Apache 2.2.17
I have
code that does a simple mysql_query(); the query on the commandline
returns an empty set. when run via PHP and the web server, the page
hangs, it never gets to the if (mysql_num_rows($result) 0) {}. and
the queries per second on mysql goes from roughly 4 per second to about
12,000.
Does anyone have any ideas?
Thanks,
Curtis
Re: [PHP] mysql problems
Does anyone have any ideas? Sounds like it's getting caught in a loop. Post the whole script for best results. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql problems
Marc Guay wrote:
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script
for
best results.
It looks like the site is
under attack, because I keep seeing the query, SELECT catagory_parent FROM
t_catagories where catagory_ID= .
$_currentCat
where $_currentCat is equal to a
value not in the database. The only way that this can happen is if
the page is called directly without going through the default page.
the script follows. its called leftNav.php
?php
include
'media/includes/productDetail.php';
//$username =
alaric;
$username = pinetree;
//$password = password_removed;
$password =
password_removed;
$hostname = 127.0.0.1;
//$hostname = www.superseeds.com;
if($_SESSION[u_id]==){
$_SESSION[u_id] = uniqid();
}
//
$_cartTotal=$0.00;
$_cartCount=0;
function tallyCart($_u_id){
global $username;
global
$password;
global $hostname;
global $_cartTotal;
global
$_cartCount;
$dbhandle =
mysql_connect($hostname, $username, $password)
or die(Unable to connect to
MySQL);
$selected =
mysql_select_db(pinetree,$dbhandle)
or die(Could not select examples);
//execute the SQL query and
return records
$result = mysql_query(SELECT
* from tbl_Cart where u_ID='.$_u_id.');
$_holder=;
$_counter=0;
$_getSubTotal=0;
$_showCheckOut=0;
while ($row = mysql_fetch_array($result)) {
$_showCheckOut=1;
$_pdetail=new
ProductDetail($row{'product_ID'}, $row{'product_Quantity'}, $_u_id);
$_getSubTotal +=
$_pdetail-_subTotal;
$_counter++;
}
$_cartTotal = $.number_format($_getSubTotal,2);
$_cartCount = $_counter;
mysql_close($dbhandle);
}
tallyCart($_SESSION[u_id]);
?
div id=div_cartCall
div id=div_cartCall_head
You have ?php echo $_cartCount? items in your
cart.br/br/
Cart total: ?php
echo $_cartTotal?
/div
div id=div_cartCall_foot
a href=cart.php#65533; Go to
cart/a
/div
/div
p
?php
//$username = alaric;
$username = pinetree;
//$password =
removed;
$password = removed;
//$hostname = 127.0.0.1;
$hostname =
www.superseeds.com;
$_parents = array();
$counter=0;
if($_GET[cat]!=){
$_parent =$_GET[cat];
}
else{
$_parent =0;
}
$dbhandle2 = mysql_connect($hostname, $username, $password)
or die(Unable to connect to MySQL);
//echo
Connected to MySQLbr;
//select a database
to work with
$selected =
mysql_select_db(pinetree,$dbhandle2)
or
die(Could not select examples);
while ($_parent !=0) {
$result_2 = mysql_query(SELECT catagory_parent
FROM t_catagories where catagory_ID= .$_parent);
$num_rows_2 =
mysql_num_rows($result_2);
if($num_rows_2 0)
{
while ($row = mysql_fetch_array($result_2)) {
$_parent= $row{'catagory_parent'};
$_parents[$counter] = $row{'catagory_parent'};
$counter++;
}
}
}
mysql_close($dbhandle2);
function getParent($catID,
$matchingID){
//$username = alaric;
$username
= pinetree;
//$password = removed;
$password = removed;
//$hostname =
127.0.0.1;
$hostname = www.superseeds.com;
$_parent=1;
$_currentCat=$catID;
$dbhandle2 = mysql_connect($hostname, $username,
$password)
or die(Unable to connect
to MySQL);
//echo Connected to
MySQLbr;
//select a database to work with
$selected =
mysql_select_db(pinetree,$dbhandle2)
or die(Could not select examples);
while
($_parent !=0) {
$result_2 = mysql_query(SELECT catagory_parent
FROM t_catagories where catagory_ID= . $_currentCat);
while
($row = mysql_fetch_array($result_2)) {
$_parent=$row{'catagory_parent'};
if($row{'catagory_parent'}==$matchingID){
mysql_close($dbhandle2);
return true;
}
}
}
mysql_close($dbhandle2);
return false;
}
?
?php
function getRowCount($_catID){
global
$_parents;
global $username;
global $password;
global
$hostname;
$dbhandle = mysql_connect($hostname, $username, $password)
or die(Unable to connect to
MySQL);
$selected = mysql_select_db(pinetree,$dbhandle)
or die(Could not select
examples);
$result = mysql_query(SELECT COUNT(*) as theCount FROM t_catagories
where catagory_parent=.$_catID);
while ($row = mysql_fetch_array($result)) {
if($row{'theCount'}==0){
mysql_close($dbhandle);
return
0;
}
else{
mysql_close($dbhandle);
return
.$row{'theCount'};
}
}
}
function
generateNav($_parent, $_style){
if(getRowCount($_parent)0){
global $_parents;
global $username;
global $password;
global $hostname;
$dbhandle3 =
Re: [PHP] mysql problems
Marc Guay wrote:
Does anyone have any ideas?
Sounds like it's getting caught in a loop. Post the whole script
for
best results.
It looks like the site is
under attack, because I keep seeing the query, SELECT catagory_parent FROM
t_catagories where catagory_ID= .
$_currentCat
where $_currentCat is equal to a
value not in the database. The only way that this can happen is if
the page is called directly without going through the default page.
the script follows. its called leftNav.php
?php
include
'media/includes/productDetail.php';
//$username =
alaric;
$username = pinetree;
//$password = password_removed;
$password =
password_removed;
$hostname = 127.0.0.1;
//$hostname = www.superseeds.com;
if($_SESSION[u_id]==){
$_SESSION[u_id] = uniqid();
}
//
$_cartTotal=$0.00;
$_cartCount=0;
function tallyCart($_u_id){
global $username;
global
$password;
global $hostname;
global $_cartTotal;
global
$_cartCount;
$dbhandle =
mysql_connect($hostname, $username, $password)
or die(Unable to connect to
MySQL);
$selected =
mysql_select_db(pinetree,$dbhandle)
or die(Could not select examples);
//execute the SQL query and
return records
$result = mysql_query(SELECT
* from tbl_Cart where u_ID='.$_u_id.');
$_holder=;
$_counter=0;
$_getSubTotal=0;
$_showCheckOut=0;
while ($row = mysql_fetch_array($result)) {
$_showCheckOut=1;
$_pdetail=new
ProductDetail($row{'product_ID'}, $row{'product_Quantity'}, $_u_id);
$_getSubTotal +=
$_pdetail-_subTotal;
$_counter++;
}
$_cartTotal = $.number_format($_getSubTotal,2);
$_cartCount = $_counter;
mysql_close($dbhandle);
}
tallyCart($_SESSION[u_id]);
?
div id=div_cartCall
div id=div_cartCall_head
You have ?php echo $_cartCount? items in your
cart.br/br/
Cart total: ?php
echo $_cartTotal?
/div
div id=div_cartCall_foot
a href=cart.php#65533; Go to
cart/a
/div
/div
p
?php
//$username = alaric;
$username = pinetree;
//$password =
removed;
$password = removed;
//$hostname = 127.0.0.1;
$hostname =
www.superseeds.com;
$_parents = array();
$counter=0;
if($_GET[cat]!=){
$_parent =$_GET[cat];
}
else{
$_parent =0;
}
$dbhandle2 = mysql_connect($hostname, $username, $password)
or die(Unable to connect to MySQL);
//echo
Connected to MySQLbr;
//select a database
to work with
$selected =
mysql_select_db(pinetree,$dbhandle2)
or
die(Could not select examples);
while ($_parent !=0) {
$result_2 = mysql_query(SELECT catagory_parent
FROM t_catagories where catagory_ID= .$_parent);
$num_rows_2 =
mysql_num_rows($result_2);
if($num_rows_2 0)
{
while ($row = mysql_fetch_array($result_2)) {
$_parent= $row{'catagory_parent'};
$_parents[$counter] = $row{'catagory_parent'};
$counter++;
}
}
}
mysql_close($dbhandle2);
function getParent($catID,
$matchingID){
//$username = alaric;
$username
= pinetree;
//$password = removed;
$password = removed;
//$hostname =
127.0.0.1;
$hostname = www.superseeds.com;
$_parent=1;
$_currentCat=$catID;
$dbhandle2 = mysql_connect($hostname, $username,
$password)
or die(Unable to connect
to MySQL);
//echo Connected to
MySQLbr;
//select a database to work with
$selected =
mysql_select_db(pinetree,$dbhandle2)
or die(Could not select examples);
while
($_parent !=0) {
$result_2 = mysql_query(SELECT catagory_parent
FROM t_catagories where catagory_ID= . $_currentCat);
while
($row = mysql_fetch_array($result_2)) {
$_parent=$row{'catagory_parent'};
if($row{'catagory_parent'}==$matchingID){
mysql_close($dbhandle2);
return true;
}
}
}
mysql_close($dbhandle2);
return false;
}
?
?php
function getRowCount($_catID){
global
$_parents;
global $username;
global $password;
global
$hostname;
$dbhandle = mysql_connect($hostname, $username, $password)
or die(Unable to connect to
MySQL);
$selected = mysql_select_db(pinetree,$dbhandle)
or die(Could not select
examples);
$result = mysql_query(SELECT COUNT(*) as theCount FROM t_catagories
where catagory_parent=.$_catID);
while ($row = mysql_fetch_array($result)) {
if($row{'theCount'}==0){
mysql_close($dbhandle);
return
0;
}
else{
mysql_close($dbhandle);
return
.$row{'theCount'};
}
}
}
function
generateNav($_parent, $_style){
if(getRowCount($_parent)0){
global $_parents;
global $username;
global $password;
global $hostname;
$dbhandle3 =
Re: [PHP] mysql problems
On Wed, May 11, 2011 at 2:25 PM, Curtis Maurand cur...@maurand.com wrote: Marc Guay wrote: Does anyone have any ideas? Sounds like it's getting caught in a loop. Post the whole script for best results. It looks like the site is under attack, because I keep seeing the query, SELECT catagory_parent FROM t_catagories where catagory_ID= . $_currentCat where $_currentCat is equal to a value not in the database. The only way that this can happen is if the page is called directly without going through the default page. the script follows. its called leftNav.php [MASSIVE SNIP] Well, from what I saw while wading through your code, you allow unsanitized variables to be concatenated to your queries. Big no-no! For ANY client-generated variable, always sanitize with mysql_real_escape_string. In fact, sanitize all your variables. It can't hurt. Also, please don't take a request for your entire code too literally. We don't like to see pages and pages and pages of code, just the pertinent bits. -- --Zootboy Sent from my PC.
[PHP] mysql problems
Hi all - I could use a lead on a problem. I just don't know where to start. I have a PHP script that populates a database table. No big deal. It creates mailing labels. However, a weird things keeps happening - every once in a while, a query is run twice. It is the same query, same information, even the same time (there is a now() in the query - and it is identical). So the question is a simple one - is this a PHP problem or a MySQL problem? Or somewhere in the MySQL extension? And how would I know? There is one clue to this otherwise vague problem. I believe that this predominantly happens when the database is under an above average load. I would appreciate any help that I might be able to get. Thank you. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql problems
Jed R. Brubaker wrote: I could use a lead on a problem. I just don't know where to start. I have a PHP script that populates a database table. No big deal. It creates mailing labels. However, a weird things keeps happening - every once in a while, a query is run twice. It is the same query, same information, even the same time (there is a now() in the query - and it is identical). So the question is a simple one - is this a PHP problem or a MySQL problem? Or somewhere in the MySQL extension? And how would I know? There is one clue to this otherwise vague problem. I believe that this predominantly happens when the database is under an above average load. I would appreciate any help that I might be able to get. Possibly the users are clicking twice on Submit when the site is slow. Try embedding an MD5 hash or some other random token in the FORM, and mark them off as used when you INSERT a new row. If a token is used just ignore the insert. Or, better yet, check that the values are the same and ignore it, and if they are different, because the user used their Back button and submitted new data, go ahead and INSERT. It might *NOT* be users clicking too much, but that's USUALLY the cause, and it's easy to detect and do the right thing once you embed something in each FORM to uniquely identify it before they submit it back to you. -- Like Music? http://l-i-e.com/artists.htm -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql problems
Jed R. Brubaker wrote: Hi all - I could use a lead on a problem. I just don't know where to start. I have a PHP script that populates a database table. No big deal. It creates mailing labels. However, a weird things keeps happening - every once in a while, a query is run twice. It is the same query, same information, even the same time (there is a now() in the query - and it is identical). So the question is a simple one - is this a PHP problem or a MySQL problem? Or somewhere in the MySQL extension? And how would I know? There is one clue to this otherwise vague problem. I believe that this predominantly happens when the database is under an above average load. I would appreciate any help that I might be able to get. Thank you. Hi Jed Can you post some sample code - showing how the query is called through PHP? IMHO, at first pass it sounds like PHP rather than MySQL - but I have been wrong before (many, many times);). Rory -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] mysql problems, need help quick
I'm filling a database with info, and I'm using id as identifiers in the list, the id is auto increment, and I deleted one entry, now I have a hole in the database, is there any way to fix this? lets say I have 1 2 3 4 5 6 And I deleted 6 Now it looks like 1 2 3 4 5 7 8 9 and so on is there somewhere the next number is located ? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] mysql problems, need help quick
Hi Hawk, (snip auto-incrementing PKs) is there somewhere the next number is located ? No. Why would you care, anyway? The thing about PKs is that they have to be unique, not sequential. If you're *really* bothered by it, you'll have to dump the contents of the table to a file, drop the table, recreate it, and import the contents into the new table. I'd also suggest, if these numbers have to be sequential, that you look hard at your db design to see if this is the right column to use as a PK. Incidentally, this is a PHP list. If you need to ask MySQL questions, please join one of the lists at http://mysql.com/. Cheers Jon -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql problems, need quick help
Primary Key's, by nature, are designed to always be uniqiue, which means that even if you delete row 6, the next row you insert will be 10 because there is no id 10. If you simply need to get the list of items in a query, in the order they were inserted, I would suggest using something like: SELECT * FROM table_name ORDER BY id ASC; Which will select everything from table_name (with no criteria such as a WHERE clause) and order it by the id in ascending order (ie, 1,2,3,4,etc.). Avid SQL users will be quick to point out that ASC is not required as ascending is the default pattern in an order by, but I still like to specify. =) Let me know if you have any other questions. Adam Voigt [EMAIL PROTECTED] On Tue, 2002-06-11 at 08:52, Hawk wrote: I'm filling a database with info, and I'm using id as identifiers in the list, the id is auto increment, and I deleted one entry, now I have a hole in the database, is there any way to fix this? lets say I have 1 2 3 4 5 6 And I deleted 6 Now it looks like 1 2 3 4 5 7 8 9 and so on is there somewhere the next number is located ? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] mysql problems
Hi on this code:
$link = mysql_connect(localhost, login, passwd);
mysql_select_db(table);
$result = mysql_query(select * from table);
while ($row = mysql_fetch_object($result)) {
echo $row-ID;
echo $row-Drank;
}
mysql_free_result($result);
mysql_close($link);
It resulst in
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 13
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 17
line 13 = while ($row = mysql_fetch_object($result)) {
line 17 = mysql_free_result($result);
I'm running an Apache/1.3.23 on a win 2000 with PHP Version 4.1.2 and mysql
3.23.39.
Does it not recoginze this mysql_... statements?
Maarten
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Re: [PHP] mysql problems
Maarten,
Perhaps table isn't the name of the table you want.
If mysql can't find the table (in line 13?), your $result variable is empty
and this causes (line 17?) to fail also.
Hope this helps,
Hugh
- Original Message -
From: Maarten Weyn [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, March 12, 2002 5:19 PM
Subject: [PHP] mysql problems
Hi on this code:
$link = mysql_connect(localhost, login, passwd);
mysql_select_db(table);
$result = mysql_query(select * from table);
while ($row = mysql_fetch_object($result)) {
echo $row-ID;
echo $row-Drank;
}
mysql_free_result($result);
mysql_close($link);
It resulst in
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 13
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 17
line 13 = while ($row = mysql_fetch_object($result)) {
line 17 = mysql_free_result($result);
I'm running an Apache/1.3.23 on a win 2000 with PHP Version 4.1.2 and
mysql
3.23.39.
Does it not recoginze this mysql_... statements?
Maarten
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RE: [PHP] mysql problems
I agree with Hugh. I think it is always a good idea to check for return
values of any function call. So, I'd replace this line:
$result = mysql_query(select * from table);
with
$result = mysql_query(select * from table);
or die(Unable to connect to SQL server: . mysql_error());
-Teresa
-Original Message-
From: hugh danaher [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, March 12, 2002 8:27 PM
To: Maarten Weyn; php
Subject: Re: [PHP] mysql problems
Maarten,
Perhaps table isn't the name of the table you want.
If mysql can't find the table (in line 13?), your $result variable is empty
and this causes (line 17?) to fail also.
Hope this helps,
Hugh
- Original Message -
From: Maarten Weyn [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, March 12, 2002 5:19 PM
Subject: [PHP] mysql problems
Hi on this code:
$link = mysql_connect(localhost, login, passwd);
mysql_select_db(table);
$result = mysql_query(select * from table);
while ($row = mysql_fetch_object($result)) {
echo $row-ID;
echo $row-Drank;
}
mysql_free_result($result);
mysql_close($link);
It resulst in
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 13
Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\index.php on line 17
line 13 = while ($row = mysql_fetch_object($result)) {
line 17 = mysql_free_result($result);
I'm running an Apache/1.3.23 on a win 2000 with PHP Version 4.1.2 and
mysql
3.23.39.
Does it not recoginze this mysql_... statements?
Maarten
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[PHP] PHP MySQL problems, updating database..
Having a problem with this, I have a working login that supports multiple users, and I'm trying to make it possible for the users to change their own settings etc, but I get killed when trying to send the data to the MySQL database. To connect I use mysql_connect($host,$user,$pswd) and mysql_select_db($db) the retrieving data from the form works since I belive, since it showed up when I added a print $the vars but I don't know how to save it to the database.. I might be missing some line or maybe it's just because I'm a newbie or something.. ;D anyway.. I use something similiar to this $query = UPDATE users SET email='$email'; // and so on.. $result = mysql_query($query) or die (blabla); I don't get any error lines or anything, the only thing that shows is the blabla thing. I also tried INSERT INTO in the query but that didn't make any difference :p if anyone could tell me what I'm doing wrong I would be happy.. :p the best way to learn is to ask people that know :p Hawk -- PHP General 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] PHP MySQL problems, updating database..
The reason you aren't seeing any errors is because you used or die() -- to the best of my knowledge, this replaces any standard errors that PHP would normally generate. It appears that you have omitted the second argument to mysql_query(). A common mistake that I've made many times myself. You need to specify the connection parameters in the second argument. $result = mysql_query($sql, $db); // $db is mysql_connect(host, user, pass) or whatever you use You would have gotten an error message telling you this if you weren't using or die... or die can be helpful but it can also occlude valuable information. Good luck I hope this works, Erik On Thursday, January 17, 2002, at 01:18 PM, Hawk wrote: Having a problem with this, I have a working login that supports multiple users, and I'm trying to make it possible for the users to change their own settings etc, but I get killed when trying to send the data to the MySQL database. To connect I use mysql_connect($host,$user,$pswd) and mysql_select_db($db) the retrieving data from the form works since I belive, since it showed up when I added a print $the vars but I don't know how to save it to the database.. I might be missing some line or maybe it's just because I'm a newbie or something.. ;D anyway.. I use something similiar to this $query = UPDATE users SET email='$email'; // and so on.. $result = mysql_query($query) or die (blabla); I don't get any error lines or anything, the only thing that shows is the blabla thing. I also tried INSERT INTO in the query but that didn't make any difference :p if anyone could tell me what I'm doing wrong I would be happy.. :p the best way to learn is to ask people that know :p Hawk -- PHP General 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] PHP MySQL problems, updating database..
Try using mysql_error() to display the mysql error message before issuing your die(). this will give more information to troubleshoot. You may not have the right privileges set up in your database where you can update or insert into the database or table. /dkm - Original Message - From: Hawk [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 17, 2002 1:18 PM Subject: [PHP] PHP MySQL problems, updating database.. Having a problem with this, I have a working login that supports multiple users, and I'm trying to make it possible for the users to change their own settings etc, but I get killed when trying to send the data to the MySQL database. To connect I use mysql_connect($host,$user,$pswd) and mysql_select_db($db) the retrieving data from the form works since I belive, since it showed up when I added a print $the vars but I don't know how to save it to the database.. I might be missing some line or maybe it's just because I'm a newbie or something.. ;D anyway.. I use something similiar to this $query = UPDATE users SET email='$email'; // and so on.. $result = mysql_query($query) or die (blabla); I don't get any error lines or anything, the only thing that shows is the blabla thing. I also tried INSERT INTO in the query but that didn't make any difference :p if anyone could tell me what I'm doing wrong I would be happy.. :p the best way to learn is to ask people that know :p Hawk -- PHP General 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 General 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] PHP MySQL problems, updating database..
Hawk, If you have a working login, can we safely assume that there is information in the database for each user? If so, then we won't bother discussing insert statements, but concentrate on updates. We'll also assume that $user has update privileges on the database. The normal form of an update query is as you describe, I'm just going to flesh it out a bit ... $query = UPDATE users SET email='$email', phone = '$phone' WHERE user_id = '$user_id' ; echo $query; $result = mysql_query( $query) or die( mysql_errno() . : .mysql_error() ); The WHERE clause makes certain that everyone's email isn't updated with the contents of $email. The echo is useful for debugging - you may be assuming that the update statement is correct, but it turns out that a key variable is missing, contains no data, or there are single quotes around numerical fields, etc. The revised die() will print out MySQL's error code and error message, which may or may not be helpful. This may still not work, but we've more information to work on. Please let us know what happens. Cheers - Miles Thompson At 07:18 PM 1/17/2002 +0100, Hawk wrote: Having a problem with this, I have a working login that supports multiple users, and I'm trying to make it possible for the users to change their own settings etc, but I get killed when trying to send the data to the MySQL database. To connect I use mysql_connect($host,$user,$pswd) and mysql_select_db($db) the retrieving data from the form works since I belive, since it showed up when I added a print $the vars but I don't know how to save it to the database.. I might be missing some line or maybe it's just because I'm a newbie or something.. ;D anyway.. I use something similiar to this $query = UPDATE users SET email='$email'; // and so on.. $result = mysql_query($query) or die (blabla); I don't get any error lines or anything, the only thing that shows is the blabla thing. I also tried INSERT INTO in the query but that didn't make any difference :p if anyone could tell me what I'm doing wrong I would be happy.. :p the best way to learn is to ask people that know :p Hawk -- PHP General 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 General 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]
