[PHP-DB] Re: www-data file
On 8/26/2014 12:20 AM, Ethan Rosenberg wrote: Dear list - When I use fopen, the file owner and group are both www-data. How can I ensure that the owner and group will be ethan? TIA Ethan Why should ownership be a concern when you are simply opening a file? AFAIK permissions are set at the time the file is placed there and will affect the access to them from then on. If you are able to fopen the file, why do the permissions matter? If you can't then you have an entirely different problem to discuss. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: www-data file
On 08/26/2014 06:21 AM, Jim Giner wrote: On 8/26/2014 12:20 AM, Ethan Rosenberg wrote: Dear list - When I use fopen, the file owner and group are both www-data. How can I ensure that the owner and group will be ethan? TIA Ethan Why should ownership be a concern when you are simply opening a file? AFAIK permissions are set at the time the file is placed there and will affect the access to them from then on. If you are able to fopen the file, why do the permissions matter? If you can't then you have an entirely different problem to discuss. fopen() can create files if they don't exist. So if you open a file for writing, it will by default receive the user/group of the web server. If you then need to access the file as some other user, you run into this situation. (I assume this is what Ethan is referring to here.) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: www-data file
On 8/26/2014 11:26 AM, Matt Pelmear wrote: fopen() can create files if they don't exist. I should have read the manual b4 replying. :( -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: No data?
Not on my system. :(
-Original Message-
From: Tamara Temple [mailto:tamo...@gmail.com]
Sent: Friday, July 27, 2012 10:46 AM
To: Brad
Cc: 'tamouse mailing lists'; php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
Brad nyctelec...@gmail.com wrote:
Wow, this is unbeileivable. Your test script fails with the attached csv
where column `a` in blank too!
No, with your attached file, it works:
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 1
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 2
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 3
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 4
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 5
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 6
}
--
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RE: [PHP-DB] Re: No data?
Even though my permissions are 777 in my /tmp I am trying to send the file
to a tmp directory in my webroot and if I can even get it to create the file
it has the wrong name of 'Array', and then it still does not update my
database.
?php
var_dump($_FILES);
//db connection
require 'dbConnect.php';
//session file
require_once('../auth.php');
function uploadList(){
$target_path = /home/nyctelecomm.com/www/mail/upload/;
//$_FILE function, grab files
$inFile = $_FILES[file];
//Get file name
$filename = $inFile[name];
//strip the file extension
$filename = preg_replace('/\.[^.]+$/','',$filename);
var_dump($_FILES);
//grab sessionID which is memberID
$memberID = $_SESSION[SESS_MEMBER_ID];
//identify cvs, tested with MS excel in csv format
if ($inFile[type] == application/octet-stream)
{
if ($inFile[error] 0)
{
echo Return Code: . $inFile['error'] . br /;
}
else
{
dbConnect();
//move_uploaded_file($_FILES[file],$target_path);
move_uploaded_file($inFile[file][tmp_name],
$target_path .'/'.$inFile);
//chmod ($filename['file'], 0777);
mysql_select_db('mailList') or die(mysql_error());
//assign temporary name to variable
//for insert since tmp_file is the real file
//$tmp_name = $inFile[file];
$presql = CREATE TABLE IF NOT EXISTS `{$memberID}`
(id MEDIUMINT AUTO_INCREMENT PRIMARY KEY UNIQUE);
$midsql = ALTER TABLE `{$memberID}` ADD
`{$filename}` VARCHAR(60);
$sql = EOF
LOAD DATA LOCAL INFILE '{$inFile}'
INTO TABLE `{$memberID}`
FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY
'\\''
LINES TERMINATED BY \\r\\n
IGNORE 1 LINES ($filename)
EOF;
var_dump($sql);
echo '$sql';
mysql_query($presql) or die(mysql_error());
mysql_query($midsql) or die(mysql_error());
mysql_query($sql) or die(mysql_error());
var_dump($sql);
echo '$sql';
if(mysql_error())
{
echo(mysql_error());
}
else
{
print('Import of campaign emails sucessfull
into mysql table.');
}
}
}
else
{
print('Invalid file type. Please make sure it is a text
file.');
}
}
//var_dump($_FILES);
uploadList();
?
*
array(1) { [file]= array(5) { [name]= string(13) testBook1.csv
[type]= string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZVJWhW [error]= int(0) [size]= int(102) } } array(1) {
[file]= array(5) { [name]= string(13) testBook1.csv [type]=
string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZVJWhW [error]= int(0) [size]= int(102) } }
Notice: Undefined index: file in /home/nyctelecomm.com/www/mail/import.php
on line 29
string(169) LOAD DATA LOCAL INFILE 'Array' INTO TABLE `3` FIELDS
TERMINATED BY ',' OPTIONALLY ENCLOSED BY '\'' LINES TERMINATED BY \r\n
IGNORE 1 LINES (testBook1) $sqlFile 'Array' not found (Errcode: 2)
-Original Message-
From: Tamara Temple [mailto:tamo...@gmail.com]
Sent: Friday, July 27, 2012 10:46 AM
To: Brad
Cc: 'tamouse mailing lists'; php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
Brad nyctelec...@gmail.com wrote:
Wow, this is unbeileivable. Your test script fails with the attached csv
where column `a` in blank too!
No, with your attached file, it works:
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 1
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 2
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 3
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 4
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 5
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 6
}
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: No data?
On 7/27/2012 11:03 AM, Brad wrote:
Even though my permissions are 777 in my /tmp I am trying to send the file
to a tmp directory in my webroot and if I can even get it to create the file
it has the wrong name of 'Array', and then it still does not update my
database.
?php
var_dump($_FILES);
//db connection
require 'dbConnect.php';
//session file
require_once('../auth.php');
function uploadList(){
$target_path = /home/nyctelecomm.com/www/mail/upload/;
//$_FILE function, grab files
$inFile = $_FILES[file];
//Get file name
$filename = $inFile[name];
//strip the file extension
$filename = preg_replace('/\.[^.]+$/','',$filename);
var_dump($_FILES);
//grab sessionID which is memberID
$memberID = $_SESSION[SESS_MEMBER_ID];
//identify cvs, tested with MS excel in csv format
if ($inFile[type] == application/octet-stream)
{
if ($inFile[error] 0)
{
echo Return Code: . $inFile['error'] . br /;
}
else
{
dbConnect();
//move_uploaded_file($_FILES[file],$target_path);
move_uploaded_file($inFile[file][tmp_name],
$target_path .'/'.$inFile);
//chmod ($filename['file'], 0777);
mysql_select_db('mailList') or die(mysql_error());
//assign temporary name to variable
//for insert since tmp_file is the real file
//$tmp_name = $inFile[file];
$presql = CREATE TABLE IF NOT EXISTS `{$memberID}`
(id MEDIUMINT AUTO_INCREMENT PRIMARY KEY UNIQUE);
$midsql = ALTER TABLE `{$memberID}` ADD
`{$filename}` VARCHAR(60);
$sql = EOF
LOAD DATA LOCAL INFILE '{$inFile}'
INTO TABLE `{$memberID}`
FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY
'\\''
LINES TERMINATED BY \\r\\n
IGNORE 1 LINES ($filename)
EOF;
var_dump($sql);
echo '$sql';
mysql_query($presql) or die(mysql_error());
mysql_query($midsql) or die(mysql_error());
mysql_query($sql) or die(mysql_error());
var_dump($sql);
echo '$sql';
if(mysql_error())
{
echo(mysql_error());
}
else
{
print('Import of campaign emails sucessfull
into mysql table.');
}
}
}
else
{
print('Invalid file type. Please make sure it is a text
file.');
}
}
//var_dump($_FILES);
uploadList();
?
*
array(1) { [file]= array(5) { [name]= string(13) testBook1.csv
[type]= string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZVJWhW [error]= int(0) [size]= int(102) } } array(1) {
[file]= array(5) { [name]= string(13) testBook1.csv [type]=
string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZVJWhW [error]= int(0) [size]= int(102) } }
Notice: Undefined index: file in /home/nyctelecomm.com/www/mail/import.php
on line 29
string(169)LOAD DATA LOCAL INFILE 'Array' INTO TABLE `3` FIELDS
TERMINATED BY ',' OPTIONALLY ENCLOSED BY '\'' LINES TERMINATED BY \r\n
IGNORE 1 LINES (testBook1) $sqlFile 'Array' not found (Errcode: 2)
-Original Message-
From: Tamara Temple [mailto:tamo...@gmail.com]
Sent: Friday, July 27, 2012 10:46 AM
To: Brad
Cc: 'tamouse mailing lists'; php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
Brad nyctelec...@gmail.com wrote:
Wow, this is unbeileivable. Your test script fails with the attached csv
where column `a` in blank too!
No, with your attached file, it works:
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 1
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 2
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 3
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 4
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 5
}
array(2) {
[0]=
string(15) br...@yahoo.com
[1]=
string(1) 6
}
You seem to be very stubborn in your attempts to handle this problem.
All the suggestions we have provided you seem to be ignoring. Dont'
understand your mentality and I surely hope you are not exhausting your
employer's resources on this effort.
Once again
RE: [PHP-DB] Re: No data?
@ Jim - Excuse me?
If you were following the thread you would see that even an outside third
party script failed, there is something else going on here.
As far as the use of $_FILES, no kidding, but when what you suggested failed
I started trying other things.
Not listening? Try to respond clear and not ambiguous, if someone isn't
following, you're not gonna squeeze blood from a turnip.
What is up with your hostility. I am so proud of you that you do this
every day, I don't. And the intent of the list is to support people working
through the kinks till they get to the point that they can bitch and gripe
at those just starting out as you do. If you're not happy with it,
unsubscribe!
Good day
-Original Message-
From: Jim Giner [mailto:jim.gi...@albanyhandball.com]
Sent: Friday, July 27, 2012 11:32 AM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
On 7/27/2012 11:03 AM, Brad wrote:
Even though my permissions are 777 in my /tmp I am trying to send the
file to a tmp directory in my webroot and if I can even get it to
create the file it has the wrong name of 'Array', and then it still
does not update my database.
?php
var_dump($_FILES);
//db connection
require 'dbConnect.php';
//session file
require_once('../auth.php');
function uploadList(){
$target_path = /home/nyctelecomm.com/www/mail/upload/;
//$_FILE function, grab files
$inFile = $_FILES[file];
//Get file name
$filename = $inFile[name];
//strip the file extension
$filename = preg_replace('/\.[^.]+$/','',$filename);
var_dump($_FILES);
//grab sessionID which is memberID
$memberID = $_SESSION[SESS_MEMBER_ID];
//identify cvs, tested with MS excel in csv format
if ($inFile[type] == application/octet-stream)
{
if ($inFile[error] 0)
{
echo Return Code: . $inFile['error'] . br
/;
}
else
{
dbConnect();
//move_uploaded_file($_FILES[file],$target_path);
move_uploaded_file($inFile[file][tmp_name],
$target_path .'/'.$inFile);
//chmod ($filename['file'], 0777);
mysql_select_db('mailList') or
die(mysql_error());
//assign temporary name to variable
//for insert since tmp_file is the real file
//$tmp_name = $inFile[file];
$presql = CREATE TABLE IF NOT EXISTS
`{$memberID}` (id MEDIUMINT AUTO_INCREMENT PRIMARY KEY UNIQUE);
$midsql = ALTER TABLE `{$memberID}` ADD
`{$filename}` VARCHAR(60);
$sql = EOF
LOAD DATA LOCAL INFILE '{$inFile}'
INTO TABLE `{$memberID}`
FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED
BY '\\''
LINES TERMINATED BY \\r\\n
IGNORE 1 LINES ($filename)
EOF;
var_dump($sql);
echo '$sql';
mysql_query($presql) or die(mysql_error());
mysql_query($midsql) or die(mysql_error());
mysql_query($sql) or die(mysql_error());
var_dump($sql);
echo '$sql';
if(mysql_error())
{
echo(mysql_error());
}
else
{
print('Import of campaign emails
sucessfull into mysql table.');
}
}
}
else
{
print('Invalid file type. Please make sure it is a
text file.');
}
}
//var_dump($_FILES);
uploadList();
?
**
**
*
array(1) { [file]= array(5) { [name]= string(13) testBook1.csv
[type]= string(24) application/octet-stream [tmp_name]=
string(14) /tmp/phpZVJWhW [error]= int(0) [size]= int(102) } }
array(1) { [file]= array(5) { [name]= string(13) testBook1.csv
[type]=
string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZVJWhW [error]= int(0) [size]= int(102) } }
Notice: Undefined index: file in
/home/nyctelecomm.com/www/mail/import.php
on line 29
string(169) LOAD DATA LOCAL INFILE 'Array' INTO TABLE `3` FIELDS
TERMINATED BY ',' OPTIONALLY ENCLOSED BY '\'' LINES TERMINATED BY
\r\n IGNORE 1 LINES (testBook1) $sqlFile 'Array' not found
(Errcode: 2)
-Original Message-
From: Tamara Temple [mailto:tamo...@gmail.com]
Sent: Friday, July 27, 2012 10:46 AM
To: Brad
Cc
RE: [PHP-DB] Re: No data?
Tamara, THANK YOU!
Switching the id field to last worked. But why?
I'm good to go now, but just curious.
Sincerely,
Brad
?php
var_dump($_FILES);
//db connection
require 'dbConnect.php';
//session file
require_once('../auth.php');
function uploadList(){
$target_path = /home/nyctelecomm.com/www/mail/upload/;
//$_FILE function, grab files
$inFile = $_FILES[file];
//Get file name
$filename = $inFile[name];
//strip the file extension
$filename = preg_replace('/\.[^.]+$/','',$filename);
var_dump($_FILES);
//grab sessionID which is memberID
$memberID = $_SESSION[SESS_MEMBER_ID];
//identify cvs, tested with MS excel in csv format
if ($inFile[type] == application/octet-stream)
{
if ($inFile[error] 0)
{
echo Return Code: . $inFile['error'] . br /;
}
else
{
dbConnect();
//move_uploaded_file($_FILES[file],$target_path);
//chmod ($filename['file'], 0777);
mysql_select_db('mailList') or die(mysql_error());
//assign temporary name to variable
//for insert since tmp_file is the real file
$tmp_name = $inFile[tmp_name];
$presql = CREATE TABLE IF NOT EXISTS `{$memberID}`
(id MEDIUMINT AUTO_INCREMENT PRIMARY KEY UNIQUE);
$midsql = ALTER TABLE `{$memberID}` ADD
`{$filename}` VARCHAR(60) FIRST;
$sql = EOF
LOAD DATA LOCAL INFILE '{$tmp_name}'
INTO TABLE `{$memberID}`
FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY
'\\''
LINES TERMINATED BY \\r\\n
IGNORE 1 LINES ($filename)
EOF;
var_dump($sql);
echo '$sql';
mysql_query($presql) or die(mysql_error());
mysql_query($midsql) or die(mysql_error());
mysql_query($sql) or die(mysql_error());
var_dump($sql);
echo '$sql';
if(mysql_error())
{
echo(mysql_error());
}
else
{
print('Import of campaign emails sucessfull
into mysql table.');
}
}
}
else
{
print('Invalid file type. Please make sure it is a text
file.');
}
}
//var_dump($_FILES);
uploadList();
?
*output*
**
array(1) { [file]= array(5) { [name]= string(13) testBook1.csv
[type]= string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZllgLU [error]= int(0) [size]= int(102) } } array(1) {
[file]= array(5) { [name]= string(13) testBook1.csv [type]=
string(24) application/octet-stream [tmp_name]= string(14)
/tmp/phpZllgLU [error]= int(0) [size]= int(102) } } string(178)
LOAD DATA LOCAL INFILE '/tmp/phpZllgLU' INTO TABLE `3` FIELDS TERMINATED BY
',' OPTIONALLY ENCLOSED BY '\'' LINES TERMINATED BY \r\n IGNORE 1 LINES
(testBook1) $sqlstring(178) LOAD DATA LOCAL INFILE '/tmp/phpZllgLU'
INTO TABLE `3` FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY '\'' LINES
TERMINATED BY \r\n IGNORE 1 LINES (testBook1) $sqlImport of campaign
emails sucessfull into mysql table.
***database
mysql show tables;
++
| Tables_in_mailList |
++
| 3 |
++
1 row in set (0.00 sec)
mysql select * from `3`;
+--++
| testBook1| id |
+--++
| br...@yahoo.com
br...@yahoo.com
br...@yahoo.com
brads@yah | 1 |
+--++
1 row in set (0.00 sec)
-Original Message-
From: Tamara Temple [mailto:tamo...@gmail.com]
Sent: Friday, July 27, 2012 12:50 PM
To: Brad
Subject: Re: [PHP-DB] Re: No data?
try this:
http://23.20.205.51/loadtablefromcsv/
See the source at http://23.20.205.51/loadtablefromcsv/index.phps
--
Tamara Temple
aka tamouse__
Email: tamo...@gmail.com
Web: http://www.tamaratemple.com/
May you never see a stranger's face in the mirror
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: No data?
Are you any closer to getting your solution to work yet? If not, could I ask for the general concept again? From what I can remember, You want to allow a user to upload a csv file which you will then put into a sql table. Additionally your code seems to be creating a new table for each file uploaded. Are these csv files each unique, ie, do they have different data contents as in different fields of information? Or do they all supply the same information in the same format. ( Hope you get what I mean.) If each file is different and each table is going to have different column names, how are these tables then going to be processed? You'll need different handlers for each table, no? And if each file is the same then why not put the data into separate records of the same table? As I mentioned before - it sure seems like you are trying to do too many things at once. Concentrate on getting the client file onto the server with a unique name and THEN work on the next problem of putting it into a table if you must. And if I am just entirely off-base, tell me to buzz off and I will leave this thread alone. I just hate to see someone struggling for so many days on something that seems so simple to me. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: No data?
1 User uploads file
2 program reads array
3 program creates a table called $memberID.$filename if not exist
($filename is column)
4 program uploads data into table/column
I am messed up at #2 for some reason so #4 fails.
All 'key's appear to be in their correct place so I don't know.
https://gist.github.com/3184155 - code
array(1) { [file]= array(5) { [name]= string(14) emailsTest.txt
[type]= string(10) text/plain [tmp_name]= string(14)
/tmp/php7wxmEI [error]= int(0) [size]= int(61) } } array(1) {
[file]= array(5) { [name]= string(14) emailsTest.txt [type]=
string(10) text/plain [tmp_name]= string(14) /tmp/php7wxmEI
[error]= int(0) [size]= int(61) } } string(165) LOAD DATA LOCAL
INFILE '/tmp/php7wxmEI' INTO TABLE `3` FIELDS TERMINATED BY ',' OPTIONALLY
ENCLOSED BY '\'' LINES TERMINATED BY \r\n IGNORE 1 LINES $sqlDuplicate
column name 'emailsTest.txt'
mysql SHOW CREATE TABLE `3`;
+---+---
-+
| Table | Create Table
|
+---+---
-+
| 3 | CREATE TABLE `3` (
`id` mediumint(9) NOT NULL AUTO_INCREMENT,
`Array` varchar(60) DEFAULT NULL,
`emailsTest.txt` varchar(60) DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `id` (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=17 DEFAULT CHARSET=utf8 |
+---+---
-+
1 row in set (0.00 sec)
mysql SELECT * FROM `3`;
Empty set (0.00 sec)
Brad Sumrall
NYCTelecomm.com
212 444-2996
-Original Message-
From: Jim Giner [mailto:jim.gi...@albanyhandball.com]
Sent: Thursday, July 26, 2012 10:55 AM
To: php-db@lists.php.net
Subject: [PHP-DB] Re: No data?
Are you any closer to getting your solution to work yet? If not, could I
ask for the general concept again? From what I can remember, You want to
allow a user to upload a csv file which you will then put into a sql table.
Additionally your code seems to be creating a new table for each file
uploaded. Are these csv files each unique, ie, do they have different data
contents as in different fields of information? Or do they all supply the
same information in the same format. ( Hope you get what I mean.) If each
file is different and each table is going to have different column names,
how are these tables then going to be processed?
You'll need different handlers for each table, no? And if each file is the
same then why not put the data into separate records of the same table?
As I mentioned before - it sure seems like you are trying to do too many
things at once. Concentrate on getting the client file onto the server with
a unique name and THEN work on the next problem of putting it into a table
if you must.
And if I am just entirely off-base, tell me to buzz off and I will leave
this thread alone. I just hate to see someone struggling for so many days
on something that seems so simple to me.
--
PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit:
http://www.php.net/unsub.php
--
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Re: [PHP-DB] Re: No data?
Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: No data?
I don't 'have' a file. It creates a temporary file and then deletes it just as fast in /tmp/. Plus, how would I $_GET it? $_FILES is supposed to handle 'ALL' of this for me? If not, what the heck does it do? Brad Sumrall NYCTelecomm.com 212 444-2996 -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:02 PM To: Brad Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Re: No data? Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] PHP-DB] Re: No data?
I apologize, I may have been mistaken about the role of $_FILES but this - http://dev.mysql.com/doc/refman/5.0/en/load-data.html confuses me. The use of $_FILES with LOAD DATA is littered over the internet and manuals as a 1 shot deal. No hand off to another directory for storage till the next function gets to it. Straight from remote drive to database in 1 single quick shot. Yours may be a work around but I honestly want it to work correctly. I'll delete the server prior to a left field hack to accommodate a failing function. :( -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:17 PM To: Brad Subject: Re: [PHP-DB] Re: No data? you need to read up on how html uploads files and how php handles them. You cannot just use the temp file (I may be wrong here, but in general one doesn't use it.). You need to use the $_FILES info in order to get the uploaded temporary file and save it permanently somewhere on your server. THEN you can use it and when through with it, you can choose to delete it. That's how it works. The code that I gave you days ago accomplishes exactly that and it gets used regularly on my site, so I know that it works. Please try it and then use an ftp client to connect to your server and see that the newly uploaded file is there. Then you can do what you wish with it. On 7/26/2012 5:10 PM, Brad wrote: I don't 'have' a file. It creates a temporary file and then deletes it just as fast in /tmp/. Plus, how would I $_GET it? $_FILES is supposed to handle 'ALL' of this for me? If not, what the heck does it do? Brad Sumrall NYCTelecomm.com 212 444-2996 -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:02 PM To: Brad Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Re: No data? Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] PHP-DB] Re: No data?
Jim I believe your correct. Upload the file to temp move the file to your dir on your server and rename file you can also set the permissions of the file while moving it. put url to file (not temp file) in database (not the file contents) done Best, Karl On Jul 26, 2012, at 4:50 PM, Brad wrote: I apologize, I may have been mistaken about the role of $_FILES but this - http://dev.mysql.com/doc/refman/5.0/en/load-data.html confuses me. The use of $_FILES with LOAD DATA is littered over the internet and manuals as a 1 shot deal. No hand off to another directory for storage till the next function gets to it. Straight from remote drive to database in 1 single quick shot. Yours may be a work around but I honestly want it to work correctly. I'll delete the server prior to a left field hack to accommodate a failing function. :( -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:17 PM To: Brad Subject: Re: [PHP-DB] Re: No data? you need to read up on how html uploads files and how php handles them. You cannot just use the temp file (I may be wrong here, but in general one doesn't use it.). You need to use the $_FILES info in order to get the uploaded temporary file and save it permanently somewhere on your server. THEN you can use it and when through with it, you can choose to delete it. That's how it works. The code that I gave you days ago accomplishes exactly that and it gets used regularly on my site, so I know that it works. Please try it and then use an ftp client to connect to your server and see that the newly uploaded file is there. Then you can do what you wish with it. On 7/26/2012 5:10 PM, Brad wrote: I don't 'have' a file. It creates a temporary file and then deletes it just as fast in /tmp/. Plus, how would I $_GET it? $_FILES is supposed to handle 'ALL' of this for me? If not, what the heck does it do? Brad Sumrall NYCTelecomm.com 212 444-2996 -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:02 PM To: Brad Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Re: No data? Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Karl DeSaulniers Design Drumm http://designdrumm.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: No data?
Your method would work. H Contemplating. Brad -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 6:02 PM To: Brad Subject: Re: [PHP-DB] Re: No data? actually, I never heard of LOAD DATA myself. But dont' think of my approach as a work-around. It's just another path to the very same result. And easier to debug. And easier to understand to another person looking at your code down the road. And it's not slower either. Maybe a msec or two. It is certainly not to be thought of as incorrect. It's how files are uploaded every day. You're just trying to consolidate two specific operations that YOU have to accomplish into one. And so far it's taken you how many days? On 7/26/2012 5:49 PM, Brad wrote: I apologize, I may have been mistaken about the role of $_FILES but this - http://dev.mysql.com/doc/refman/5.0/en/load-data.html confuses me. The use of $_FILES with LOAD DATA is littered over the internet and manuals as a 1 shot deal. No hand off to another directory for storage till the next function gets to it. Straight from remote drive to database in 1 single quick shot. Yours may be a work around but I honestly want it to work correctly. I'll delete the server prior to a left field hack to accommodate a failing function. :( -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:17 PM To: Brad Subject: Re: [PHP-DB] Re: No data? you need to read up on how html uploads files and how php handles them. You cannot just use the temp file (I may be wrong here, but in general one doesn't use it.). You need to use the $_FILES info in order to get the uploaded temporary file and save it permanently somewhere on your server. THEN you can use it and when through with it, you can choose to delete it. That's how it works. The code that I gave you days ago accomplishes exactly that and it gets used regularly on my site, so I know that it works. Please try it and then use an ftp client to connect to your server and see that the newly uploaded file is there. Then you can do what you wish with it. On 7/26/2012 5:10 PM, Brad wrote: I don't 'have' a file. It creates a temporary file and then deletes it just as fast in /tmp/. Plus, how would I $_GET it? $_FILES is supposed to handle 'ALL' of this for me? If not, what the heck does it do? Brad Sumrall NYCTelecomm.com 212 444-2996 -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:02 PM To: Brad Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Re: No data? Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] PHP-DB] Re: No data?
Drats, I think you are correct. :) -Original Message- From: Karl DeSaulniers [mailto:k...@designdrumm.com] Sent: Thursday, July 26, 2012 5:59 PM To: php-db@lists.php.net Subject: Re: [PHP-DB] PHP-DB] Re: No data? Jim I believe your correct. Upload the file to temp move the file to your dir on your server and rename file you can also set the permissions of the file while moving it. put url to file (not temp file) in database (not the file contents) done Best, Karl On Jul 26, 2012, at 4:50 PM, Brad wrote: I apologize, I may have been mistaken about the role of $_FILES but this - http://dev.mysql.com/doc/refman/5.0/en/load-data.html confuses me. The use of $_FILES with LOAD DATA is littered over the internet and manuals as a 1 shot deal. No hand off to another directory for storage till the next function gets to it. Straight from remote drive to database in 1 single quick shot. Yours may be a work around but I honestly want it to work correctly. I'll delete the server prior to a left field hack to accommodate a failing function. :( -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:17 PM To: Brad Subject: Re: [PHP-DB] Re: No data? you need to read up on how html uploads files and how php handles them. You cannot just use the temp file (I may be wrong here, but in general one doesn't use it.). You need to use the $_FILES info in order to get the uploaded temporary file and save it permanently somewhere on your server. THEN you can use it and when through with it, you can choose to delete it. That's how it works. The code that I gave you days ago accomplishes exactly that and it gets used regularly on my site, so I know that it works. Please try it and then use an ftp client to connect to your server and see that the newly uploaded file is there. Then you can do what you wish with it. On 7/26/2012 5:10 PM, Brad wrote: I don't 'have' a file. It creates a temporary file and then deletes it just as fast in /tmp/. Plus, how would I $_GET it? $_FILES is supposed to handle 'ALL' of this for me? If not, what the heck does it do? Brad Sumrall NYCTelecomm.com 212 444-2996 -Original Message- From: Jim Giner [mailto:jim.gi...@albanyhandball.com] Sent: Thursday, July 26, 2012 5:02 PM To: Brad Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Re: No data? Not sure what you mean - Program reads array. What program is doing is utilizing the FILES element to get the info about the uploaded file and proceeds to finish the upload by moving it to a temporary folder of your creation. Once that is done you HAVE the file under whatever name you want to call it in whatever folder you want it in. Why don't you just do that part and then use some tool to look at your server structure to verify that you have the file. NOW you can proceed with the rest, if you must. You're going to create a table with a column/field name matching the filename and then you'll put the contents of that file into that single column in one record of this table. So now you have a one record table with one column holding the contents of a file. How is this different from just having the file? 1 User uploads file 2 program reads array 3 program creates a table called $memberID.$filename if not exist ($filename is column) 4 program uploads data into table/column I am messed up at #2 for some reason so #4 fails. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Karl DeSaulniers Design Drumm http://designdrumm.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: No data?
Brad, do take a look at http://pastie.org/4340383 It does work. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: No data?
With one change to the data file to make the indexing work correctly
and read all the fields: to accommodate the first id auto_increment
field, the first value in the csv line must be null:
null.a,b,c,d,e
null,f,g,h,i,j
null,k,l,m,n,o
And further, the field count should subsequently be one less in line
53 when creating the table:
for ($i=0; $i (count($fields)-1); $i++) {
and that should do it.
On Thu, Jul 26, 2012 at 10:20 PM, tamouse mailing lists
tamouse.li...@gmail.com wrote:
Brad, do take a look at http://pastie.org/4340383
It does work.
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Re: [PHP-DB] Re: No data?
argh
forget that.
simply put the id auto_increment field *last* in the create table
defn, leave everything else as it was.
On Thu, Jul 26, 2012 at 10:38 PM, tamouse mailing lists
tamouse.li...@gmail.com wrote:
With one change to the data file to make the indexing work correctly
and read all the fields: to accommodate the first id auto_increment
field, the first value in the csv line must be null:
null.a,b,c,d,e
null,f,g,h,i,j
null,k,l,m,n,o
And further, the field count should subsequently be one less in line
53 when creating the table:
for ($i=0; $i (count($fields)-1); $i++) {
and that should do it.
On Thu, Jul 26, 2012 at 10:20 PM, tamouse mailing lists
tamouse.li...@gmail.com wrote:
Brad, do take a look at http://pastie.org/4340383
It does work.
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RE: [PHP-DB] Re: No data?
No, it did not work.
mysql show tables;
++
| Tables_in_mailList |
++
| testBook1csv |
++
1 row in set (0.00 sec)
mysql select * from `testBook1csv`;
+++
| id | field0 |
+++
| 1 | NULL |
| 2 | NULL |
| 3 | NULL |
| 4 | NULL |
| 5 | NULL |
| 6 | NULL |
+++
6 rows in set (0.00 sec)
file upload example
$_FILES
array(1) {
[userfile]=
array(5) {
[name]=
string(13) testBook1.csv
[type]=
string(24) application/octet-stream
[tmp_name]=
string(14) /tmp/phpiaW23p
[error]=
int(0)
[size]=
int(102)
}
}
ls -l /tmp/phpiaW23p
-rw--- 1 apache apache 102 Jul 27 00:02 /tmp/phpiaW23p
$sql for create table
string(110) create table if not exists `testBook1csv` (id integer not null
auto_increment primary key, field0 varchar(50))
$sql for load data
string(142) load data local infile '/tmp/phpiaW23p' into table `testBook1csv`
fields terminated by ',' optionally enclosed by '' lines terminated by '\n'
Verify data contents
Table testBook1csv
string(28) select * from `testBook1csv`
$sql=
data
array(2) {
[0]=
string(1) 1
[1]=
NULL
}
array(2) {
[0]=
string(1) 2
[1]=
NULL
}
array(2) {
[0]=
string(1) 3
[1]=
NULL
}
array(2) {
[0]=
string(1) 4
[1]=
NULL
}
array(2) {
[0]=
string(1) 5
[1]=
NULL
}
array(2) {
[0]=
string(1) 6
[1]=
NULL
}
-Original Message-
From: tamouse mailing lists [mailto:tamouse.li...@gmail.com]
Sent: Thursday, July 26, 2012 11:20 PM
To: Brad
Cc: jim.gi...@albanyhandball.com; php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
Brad, do take a look at http://pastie.org/4340383 It does work.
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RE: [PHP-DB] Re: No data?
Wow, this is unbeileivable. Your test script fails with the attached csv where
column `a` in blank too!
Php.ini
http://pastie.org/4340845
mysql select * from `testBook1csv`;
+++
| id | field0 |
+++
| 1 | NULL |
| 2 | NULL |
| 3 | NULL |
| 4 | NULL |
| 5 | NULL |
| 6 | NULL |
+++
6 rows in set (0.00 sec)
*output from your script**
file upload example
$_FILES
array(1) {
[userfile]=
array(5) {
[name]=
string(13) testBook1.csv
[type]=
string(24) application/octet-stream
[tmp_name]=
string(14) /tmp/phphf42G8
[error]=
int(0)
[size]=
int(102)
}
}
ls -l /tmp/phphf42G8
-rw--- 1 apache apache 102 Jul 27 00:17 /tmp/phphf42G8
$sql for create table
string(110) create table if not exists `testBook1csv` (id integer not null
auto_increment primary key, field0 varchar(50))
$sql for load data
string(142) load data local infile '/tmp/phphf42G8' into table `testBook1csv`
fields terminated by ',' optionally enclosed by '' lines terminated by '\n'
Verify data contents
Table testBook1csv
string(28) select * from `testBook1csv`
$sql=
data
array(2) {
[0]=
string(1) 1
[1]=
NULL
}
array(2) {
[0]=
string(1) 2
[1]=
NULL
}
array(2) {
[0]=
string(1) 3
[1]=
NULL
}
array(2) {
[0]=
string(1) 4
[1]=
NULL
}
array(2) {
[0]=
string(1) 5
[1]=
NULL
}
array(2) {
[0]=
string(1) 6
[1]=
NULL
}
-Original Message-
From: tamouse mailing lists [mailto:tamouse.li...@gmail.com]
Sent: Thursday, July 26, 2012 11:39 PM
To: Brad
Cc: jim.gi...@albanyhandball.com; php-db@lists.php.net
Subject: Re: [PHP-DB] Re: No data?
With one change to the data file to make the indexing work correctly and read
all the fields: to accommodate the first id auto_increment field, the first
value in the csv line must be null:
null.a,b,c,d,e
null,f,g,h,i,j
null,k,l,m,n,o
And further, the field count should subsequently be one less in line
53 when creating the table:
for ($i=0; $i (count($fields)-1); $i++) {
and that should do it.
On Thu, Jul 26, 2012 at 10:20 PM, tamouse mailing lists
tamouse.li...@gmail.com wrote:
Brad, do take a look at http://pastie.org/4340383 It does work.
testBook1.csv
Description: MS-Excel spreadsheet
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[PHP-DB] Re: Extracting data from Arrays in ISAM table
Step #1: Select name_field FROM the_table;
Step #2:
$arr_all_names = array();
foreach single_result in your result_set{ //you have to translate this into
php
$arr_name = explode(',', $single_result['name']); //assuming that you
are joining names in the field using a comma
$arr_all_names = $arr_all_names + $arr_name;
}
//remove redundancy
$arr_all_names = array_unique($arr_all_names);
//now sort the array
sort($arr_all_names);
//now all you have to do is to loop through the array to display it on your
site
--
itoctopus - http://www.itoctopus.com
[EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED]
I have a table with between 100k and 200k rows. One field, `names`, is
populated, in each row, with an imploded array of up to 4 names.
I require to create a list of names, without repeats, of all the names
in `names` field to use in an html form.
Any advise would be appreciated. I have no ideas on how to start.
Louise
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[PHP-DB] Re: Transfer data between tables in MySQL
Rosen wrote: Hi, I have to transfer all data between two tables ( with identical structure ) Is this possible with one query, or I must read from table1 and manually insert into table2? Thanks in advance, Rosen INSERT INTO `dbase2`.`table1` SELECT * FROM `dbase1`.`table1` ; That will copy from 1st database to second. If you use phpMyAdmin you can do it via the operations tab. BMA -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Highlighting data selected from one table that appear in another
I'm thinking that you need a LEFT JOIN (see the mysql documentation) to do this. Basically LEFT JOIN (or RIGHT JOIN) allow you to select all the rows from the table on the left side even if they have no corresponding row on the right side (the values of the columns in the select that belong to the right hand table will be NULL where there is no row to join to). But if you could be more specific - send the schema and precisely what you were looking to do I could give you an example of how to make it work. Good Luck, Frank On Sep 1, 2005, at 5:00 AM, [EMAIL PROTECTED] wrote: From: boclair [EMAIL PROTECTED] Date: September 1, 2005 4:50:05 AM PDT To: php-db@lists.php.net php-db@lists.php.net Subject: Highlighting data selected from one table that appear in another Content-Type: multipart/alternative; boundary=070700030402040702060708 X-Authentication-Info: Submitted using SMTP AUTH PLAIN at omta05ps.mx.bigpond.com from [138.130.220.111] using ID boclair at Thu, 1 Sep 2005 11:49:59 + This is a multi-part message in MIME format. --070700030402040702060708 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit mysql 4.0.25 php 4.3.1 I seem to be unable to solve this problem without some help. A webpage is to list all the clients in a mysql Client table. Each client has an an allocated clid. Those clients who are listed, identified by their allocated clid, in a Transactions are to be highlighted in different ways according to progress of the transaction, the progress being determined on whether date values have been inserted into various fields. Since mysql select joining the tables lists only the clients whose appear in both tables. I have been attempting to create a temporary table inserting the data from Client and Transactions and selecting from that but not successfully yet. But is this the way to go about it. The scale is an anticipated 1000 clients per month of which an anticipated 800 will involve transactions Louise
[PHP-DB] Re: Highlighting data selected from one table that appear in another
Frank Flynn wrote: But if you could be more specific - send the schema and precisely what you were looking to do I could give you an example of how to make it work. I am still having difficulties with my workup and more help would be appreciated . I have two tables CREATE TABLE `clients` ( `id` tinyint(4) NOT NULL auto_increment, `clid` varchar(5) NOT NULL default '', `clfname` varchar(50) NOT NULL default '', `cllname` varchar(30) NOT NULL default '', PRIMARY KEY (`id`), UNIQUE KEY `clid` (`clid`) ) TYPE=MyISAM COMMENT='workup table for client data'; CREATE TABLE `transactions` ( `id` tinyint(4) NOT NULL auto_increment, `fmid` varchar(5) NOT NULL default '', `clid` varchar(5) NOT NULL default '', `bdate` date NOT NULL default '-00-00', `instr` mediumtext, `ecom` decimal(5,2) default NULL, `efdate` date default '-00-00', `acom` decimal(5,2) default NULL, `afdate` date default '-00-00', `transid` varchar(10) NOT NULL default '', PRIMARY KEY (`id`,`fmid`), UNIQUE KEY `transid` (`transid`), KEY `fmid` (`fmid`) ) TYPE=MyISAM; COMMENT='workup table for transaction data'; What is needed is to display a list of all clients each denoted applicably as 1.. No entry in transaction table (client's clid not in transaction table) 2.. Client not yet interviewed ( instr empty, efdate default value and afdate default value) 3.. Client interviewed ( instr !empty, efdate=default value and afdate=default value) 4.. Client's transaction in progress ( instr !empty, efdate!=default value and afdate=default value) 5.. Clients transaction finalised ( instr !empty, efdate!=default value and afdate!=default value) What I appear to be getting is LEFT JOIN only selects those clients with a matching clid in both tables LEFT OUTER JOIN selects all clients with the appropriate transaction.fmid (for example) except when the client.clid is not matched by a transaction.clid when a spurious fmid is given. SELECT * FROM clients LEFT OUTER JOIN transaction ON clients.clid = transactions.clid ORDER BY clients.cllname ASC I am obviously handling the join incorrectly. Louise -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: load data infile -- problem
select.now [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] So, in command line, all works fine, all records are correctly imported. command line code - mysql load data infile 'c:\\datastream\\import\import -into table iport - fields terminated by ';' - ignore 9 lines; - /command line code -- php code --- $a = 'c:\\\datastream\\\import\\\import'; $extra_arg = 'fields terminated by \';\' ignore 9 lines;'; $query = 'load data infile \''.$a.'\' into table import '.$extra_arg.''; $result = mysql_query ($query) or die_mysql (brExecutia comenzii i$query/i a esuat.br); -- /php code -- I get this output-error: === load data infile 'c:\\datastream\\import\\import' into table import fields terminated by ';' ignore 9 lines; === It looks like your slashes are a wee-bit messed up. Maybe try: $a = 'c:\\datastream\\import\\import'; Note: a single '\' means to escape the next character, so two '\\' means to put in a single '\'. You had three '\\\'. DanB -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Moving data from one MySQL table to another
First off - is this a solution in search of a problem? Databases are excellent at indexing and caching frequently used data. Though your table has millions of rows and you are only really interested in a few hundred you might find that performance is not noticeably any different than if you only had a few hundred rows total. And if there is a big difference look at the indexes - really, it seems unbelievable but in fact it is true. But if there is a big difference you can move the data with these two statements (and the data does not ever leave the DB). INSERT INTO newTable (column1, column2, ...) SELECT column1, column2, ... FROM oldTable WHERE column3 some date /* or whatever */ DELETE FROM oldTable WHERE column3 some date /* or whatever */ Be sure the two where clauses match exactly. Good Luck, Frank On May 19, 2005, at 3:28 PM, [EMAIL PROTECTED] wrote: From: Jeffrey [EMAIL PROTECTED] Date: May 19, 2005 6:19:48 AM PDT To: php-db@lists.php.net Subject: Moving data from one MySQL table to another I'm working on a web application and one of the things I am doing is creating an archiving function that would move older data to archive tables in order to minimise the amount of data in the active tables. This so that the data that is being used more frequently can be accessed faster by the users. My approach in building the archive function is: 1) SELECT query on the data 2) mysql_fetch_array to put the data into an array 3) INSERT subqueries to put the data into the archive tables. My concern is that in some cases, hundreds of rows of data would need to be moved - which could lead to awfully big arrays. However, the archiving function is likely to be used infrequently - not more than 1 or 2 times per week. This leads to two questions: 1) Could such a big array cause performance problems or worse? 2) Is there a better way? Many thanks, Jeff
[PHP-DB] Re: MySQL Data Not Returned
My guess (and it will be correct) is that you've got a not visible character on the end of your email address in the database. Such a character could be a space, newline, tab or other similar character. Make sure you use trim($email_address) to create the field to insert into the database ! Check the actual string you get returned from the query on the DB with the query SELECT * FROM mailinglist WHERE email like '$email%' Use strlen($resul[email]); and you'll see that the length of the email address is not the same as the number of characters in the email address if you count them manually. Cheers - Neil At 11:54 13/11/2004 +, you wrote: I search the list like so: $email=[EMAIL PROTECTED]; $query=mysqli_query($cnn,SELECT * FROM mailinglist WHERE email='$email'); When I do: Echo mysqli_num_rows($query); I get 0 back, The address I am searching is in the database, I have triple checked it with phpmyadmin. The only way I can get it to return the record I am looking for is to do this. $email=[EMAIL PROTECTED]; $query=mysqli_query($cnn,SELECT * FROM mailinglist WHERE email like '$email%'); Then this finally returns 1. I have tried many different email addresses; nothing comes back, not even from the SQL query window within phpmyadmin v. 2.6.0-rc2 This has been puzzling me for hours on end. If anyone can provide any help, that would be great -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: LOAD DATA LOCAL INFILE
Hi Dani et al.. The guys at MySQL turned this off for security reasons. :( They and SuSE say that you can turn it back on, but I've never been able to get it to work. There are two ways to handle this, first you have to use LOAD DATA INFILE. The file then needs to be either in /tmp and readable by everyone (this is best). Or the way I do it is put in the same dir where the database is, ie /var/lib/mysql/databaseXXX. The problem with this is that you have to change the premissions. Which is not a good idea on an internet machine. Hope this helps, JIM On Monday 16 August 2004 14:40, Daniel Kradolfer wrote: Hello I'm new to this list. I diden't found anything in the archive and by googleing to solve my problem. We have updated our server to PHP 4.3.8 and MySQL 4.0.2. Some of our clients run a shop software that need 'LOAD DATA LOCAL INFILE' to store its data to MySQL. MySQL itself supports the command. I can run it from mysql-client. That works. I have read about the fix on php.net (http://www.php.net/ChangeLog-4.php / http://bugs.php.net/28632). Until now, i couldn't find a solution to get it work. Is there way to get it work? Regards Dani -- Jim Hatridge Linux User #88484 -- BayerWulf Linux System # 129656 The Recycled Beowulf Project Looking for throw-away or obsolete computers and parts to recycle into a Linux super computer WartHog Bulletin Info about new German Stamps http://www.fuzzybunnymilitia.org/~hatridge/bulletin Viel Feind -- Viel Ehr' Anti-US Propaganda stamp collection http://www.fuzzybunnymilitia.org/~hatridge/collection -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: LOAD DATA LOCAL INFILE
Am 2004-08-16 19:11:50, schrieb James Hatridge: Hi Dani et al.. The guys at MySQL turned this off for security reasons. :( They and SuSE say that you can turn it back on, but I've never been able to get it to work. There are two ways to handle this, first you have to use LOAD DATA INFILE. The file then needs to be either in /tmp and readable by everyone (this is best). Or the way I do it is put in the same dir where the database is, ie /var/lib/mysql/databaseXXX. The problem with this is that you have to change the premissions. Which is not a good idea on an internet machine. Hope this helps, What about: http://www.ispirer.com/doc/sqlways36/troubleshooting/mysql_db.html In MySQL 4.0.x LOCAL will only work if MySQL server was started with --local-infile=1. JIM Hello, Greetings Michelle -- Linux-User #280138 with the Linux Counter, http://counter.li.org/ Michelle Konzack Apt. 917 ICQ #328449886 50, rue de Soultz MSM LinuxMichi 0033/3/8845235667100 Strasbourg/France IRC #Debian (irc.icq.com) signature.pgp Description: Digital signature
[PHP-DB] Re: Exporting Data From MySQL Using PHP
Hello Ron, If I understand correctly, you're asking what query to execute (and how to put the result in a textfile) to get the results you want? The first part is described thorougly in the mysql-manual (chapter 14.1.7 for example, search for the SELECT syntax or examples. You'll find it at http://dev.mysql.com/doc/mysql/en/ ). You probably need a query that looks like this: SELECT e_mail FROM your_table_name WHERE discipleship_mailing_list_e_mail_subscription = on; This will return all the requested e-mail adresses. Having this information printed to a file could be reached by using the INTO OUTFILE form of SELECT. Again, check the mysql-manual. for the exact syntax, but you'll probably end up with somthing like: SELECT e_mail INTO OUTFILE '/some/filename.text' FROM your_table_name WHERE discipleship_mailing_list_e_mail_subscription = on; There are a bunch of other options and solutions, though this seems the easiest one to me. Mind that if you try to do this in PHP (and trough a webserver), the webserver should have write-access to the file you specify with 'OUTFILE' - that might be a problem. Personally, I'd run this as some sort of batch script (using the mysql command line interface with the -e option, for example). Hope this helps you along a bit. Regards, Guus der Kinderen Ron Piggott wrote: I have created a MySQL database. The table I am creating is a subscriptions database. I want to be able to export all e-mail addresses stored in the e_mail column into a plain text file on the web server (1 e-mail address per row) where the discipleship_mailing_list_e_mail_subscription equals on. I am new at PHP and I am not sure how to do this yet. My idea is that the user would click an UPDATE button and this action would be performed. Are any of you able to help me with this? Thanks. Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Exporting Data From MySQL Using PHP
It is a problem having write access to the file. This is what the online mySQL manual says ... The SELECT ... INTO OUTFILE 'file_name' form of SELECT writes the selected rows to a file. The file is created on the server host, so you must have the FILE privilege to use this syntax. The file cannot already exist, which among other things prevents files such as `/etc/passwd' and database tables from being destroyed. The SELECT ... INTO OUTFILE statement is intended primarily to let you very quickly dump a table on the server machine. If you want to create the resulting file on some client host other than the server host, you can't use SELECT ... INTO OUTFILE. In that case, you should instead use some command like mysql -e SELECT ... file_name on the client host to generate the file. I need to go with another option. Another person wrote me and suggested using the fopen(), fwrite() and fclose() commands. I haven't investigated this option yet. You know when you first learn to walk you do so by example ... I am still needing to know which commands to look up and try to apply to my situation. I have got some PHP code to work ... I am far from done the project I am working on. I was able to get the command: SELECT e_mail FROM subscriptionsdatabase WHERE discipleship_mailing_list_e_mail_subscription LIKE 'on' to work in the mySQL command prompt but when I put it into a .PHP file I get a parse error. I get that parse error by simply coping and pasting the command into the PHP file. In more detail what I am trying to do and why I e-mailed out the question is to wipe clean a mailing list data file and re-create the e-mail subscriptions data file based on which people joined the list during the week. I am still learning mySQL ... The things you wrote helped me play ... I didn't get too far in creating code, but I am learning at the same time. Thanks for the advice. Ron - Original Message - From: Guus der Kinderen [EMAIL PROTECTED] Newsgroups: php.db To: Ron Piggott [EMAIL PROTECTED] Sent: Monday, August 09, 2004 7:22 PM Subject: Re: Exporting Data From MySQL Using PHP Hello Ron, If I understand correctly, you're asking what query to execute (and how to put the result in a textfile) to get the results you want? The first part is described thorougly in the mysql-manual (chapter 14.1.7 for example, search for the SELECT syntax or examples. You'll find it at http://dev.mysql.com/doc/mysql/en/ ). You probably need a query that looks like this: SELECT e_mail FROM your_table_name WHERE discipleship_mailing_list_e_mail_subscription = on; This will return all the requested e-mail adresses. Having this information printed to a file could be reached by using the INTO OUTFILE form of SELECT. Again, check the mysql-manual. for the exact syntax, but you'll probably end up with somthing like: SELECT e_mail INTO OUTFILE '/some/filename.text' FROM your_table_name WHERE discipleship_mailing_list_e_mail_subscription = on; There are a bunch of other options and solutions, though this seems the easiest one to me. Mind that if you try to do this in PHP (and trough a webserver), the webserver should have write-access to the file you specify with 'OUTFILE' - that might be a problem. Personally, I'd run this as some sort of batch script (using the mysql command line interface with the -e option, for example). Hope this helps you along a bit. Regards, Guus der Kinderen Ron Piggott wrote: I have created a MySQL database. The table I am creating is a subscriptions database. I want to be able to export all e-mail addresses stored in the e_mail column into a plain text file on the web server (1 e-mail address per row) where the discipleship_mailing_list_e_mail_subscription equals on. I am new at PHP and I am not sure how to do this yet. My idea is that the user would click an UPDATE button and this action would be performed. Are any of you able to help me with this? Thanks. Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Exporting Data From MySQL Using PHP
It has been my experience that you sometimes need to do backquotes (`)
in MySQL queries when cutting/pasting into PHP -- it's the same key as
the tilde (~) on my keyboard. Try:
$result = mysql_query(SELECT e_mail FROM subscriptionsdatabase WHERE
`discipleship_mailing_list_e_mail_subscription` LIKE 'on');
That should work. I've always used the equal sign instead of LIKE, but
that seems to be a matter of preference.
Regarding the second part, with the fopen() and fwrite() commands,
that's certainly doable. For example, you could do something like:
$handle = fopen(/home/blah/blah.txt, w);
$result = mysql_query(...);
if($result) {
while($somevar = do mysql fetch from $result) {
fwrite($handle, $somevar);
}
...
}
fclose($result);
Now, I realize that code is sort of a hack, but since I'm doing it
mostly on the fly, you'll forgive me my coding errors. Take a look at
the Filesystem section of the PHP manual for information on the
fopen/fwrite() commands, and the MySQL section for more information on
MySQL commands in PHP.
Hope this helps a little more in solving the puzzle.
--
Peter Ellis - [EMAIL PROTECTED]
Web Design and Development Consultant
naturalaxis | http://www.naturalaxis.com/
On Mon, 2004-08-09 at 22:44 -0400, Ron Piggott wrote:
It is a problem having write access to the file. This is what the online
mySQL manual says ...
The SELECT ... INTO OUTFILE 'file_name' form of SELECT writes the selected
rows to a file. The file is created on the server host, so you must have the
FILE privilege to use this syntax. The file cannot already exist, which
among other things prevents files such as `/etc/passwd' and database tables
from being destroyed. The SELECT ... INTO OUTFILE statement is intended
primarily to let you very quickly dump a table on the server machine. If you
want to create the resulting file on some client host other than the server
host, you can't use SELECT ... INTO OUTFILE. In that case, you should
instead use some command like mysql -e SELECT ... file_name on the
client host to generate the file.
I need to go with another option.
Another person wrote me and suggested using the fopen(), fwrite() and
fclose() commands. I haven't investigated this option yet. You know when
you first learn to walk you do so by example ... I am still needing to know
which commands to look up and try to apply to my situation. I have got some
PHP code to work ... I am far from done the project I am working on.
I was able to get the command:
SELECT e_mail
FROM subscriptionsdatabase
WHERE discipleship_mailing_list_e_mail_subscription
LIKE 'on'
to work in the mySQL command prompt but when I put it into a .PHP file I get
a parse error. I get that parse error by simply coping and pasting the
command into the PHP file.
In more detail what I am trying to do and why I e-mailed out the question is
to wipe clean a mailing list data file and re-create the e-mail
subscriptions data file based on which people joined the list during the
week.
I am still learning mySQL ... The things you wrote helped me play ... I
didn't get too far in creating code, but I am learning at the same time.
Thanks for the advice.
Ron
- Original Message -
From: Guus der Kinderen [EMAIL PROTECTED]
Newsgroups: php.db
To: Ron Piggott [EMAIL PROTECTED]
Sent: Monday, August 09, 2004 7:22 PM
Subject: Re: Exporting Data From MySQL Using PHP
Hello Ron,
If I understand correctly, you're asking what query to execute (and how
to put the result in a textfile) to get the results you want?
The first part is described thorougly in the mysql-manual (chapter
14.1.7 for example, search for the SELECT syntax or examples. You'll
find it at http://dev.mysql.com/doc/mysql/en/ ). You probably need a
query that looks like this:
SELECT e_mail FROM your_table_name
WHERE discipleship_mailing_list_e_mail_subscription = on;
This will return all the requested e-mail adresses.
Having this information printed to a file could be reached by using the
INTO OUTFILE form of SELECT. Again, check the mysql-manual. for the
exact syntax, but you'll probably end up with somthing like:
SELECT e_mail INTO OUTFILE '/some/filename.text' FROM your_table_name
WHERE discipleship_mailing_list_e_mail_subscription = on;
There are a bunch of other options and solutions, though this seems the
easiest one to me. Mind that if you try to do this in PHP (and trough a
webserver), the webserver should have write-access to the file you
specify with 'OUTFILE' - that might be a problem. Personally, I'd run
this as some sort of batch script (using the mysql command line
interface with the -e option, for example).
Hope this helps you along a bit.
Regards,
Guus der Kinderen
Ron Piggott wrote:
I have created a MySQL database. The table I am creating is a
subscriptions
database. I want to be able to export all e-mail addresses stored
Re: [PHP-DB] Re: Exporting Data From MySQL Using PHP
Whoops -- the fclose() statement refers to $result when it should refer
to $handle. My bad for not at least spot checking the message before
hitting send!
--
Peter Ellis - [EMAIL PROTECTED]
Web Design and Development Consultant
naturalaxis | http://www.naturalaxis.com/
On Mon, 2004-08-09 at 20:34 -0700, Peter Ellis wrote:
It has been my experience that you sometimes need to do backquotes (`)
in MySQL queries when cutting/pasting into PHP -- it's the same key as
the tilde (~) on my keyboard. Try:
$result = mysql_query(SELECT e_mail FROM subscriptionsdatabase WHERE
`discipleship_mailing_list_e_mail_subscription` LIKE 'on');
That should work. I've always used the equal sign instead of LIKE, but
that seems to be a matter of preference.
Regarding the second part, with the fopen() and fwrite() commands,
that's certainly doable. For example, you could do something like:
$handle = fopen(/home/blah/blah.txt, w);
$result = mysql_query(...);
if($result) {
while($somevar = do mysql fetch from $result) {
fwrite($handle, $somevar);
}
...
}
fclose($result);
Now, I realize that code is sort of a hack, but since I'm doing it
mostly on the fly, you'll forgive me my coding errors. Take a look at
the Filesystem section of the PHP manual for information on the
fopen/fwrite() commands, and the MySQL section for more information on
MySQL commands in PHP.
Hope this helps a little more in solving the puzzle.
--
Peter Ellis - [EMAIL PROTECTED]
Web Design and Development Consultant
naturalaxis | http://www.naturalaxis.com/
On Mon, 2004-08-09 at 22:44 -0400, Ron Piggott wrote:
It is a problem having write access to the file. This is what the online
mySQL manual says ...
The SELECT ... INTO OUTFILE 'file_name' form of SELECT writes the selected
rows to a file. The file is created on the server host, so you must have the
FILE privilege to use this syntax. The file cannot already exist, which
among other things prevents files such as `/etc/passwd' and database tables
from being destroyed. The SELECT ... INTO OUTFILE statement is intended
primarily to let you very quickly dump a table on the server machine. If you
want to create the resulting file on some client host other than the server
host, you can't use SELECT ... INTO OUTFILE. In that case, you should
instead use some command like mysql -e SELECT ... file_name on the
client host to generate the file.
I need to go with another option.
Another person wrote me and suggested using the fopen(), fwrite() and
fclose() commands. I haven't investigated this option yet. You know when
you first learn to walk you do so by example ... I am still needing to know
which commands to look up and try to apply to my situation. I have got some
PHP code to work ... I am far from done the project I am working on.
I was able to get the command:
SELECT e_mail
FROM subscriptionsdatabase
WHERE discipleship_mailing_list_e_mail_subscription
LIKE 'on'
to work in the mySQL command prompt but when I put it into a .PHP file I get
a parse error. I get that parse error by simply coping and pasting the
command into the PHP file.
In more detail what I am trying to do and why I e-mailed out the question is
to wipe clean a mailing list data file and re-create the e-mail
subscriptions data file based on which people joined the list during the
week.
I am still learning mySQL ... The things you wrote helped me play ... I
didn't get too far in creating code, but I am learning at the same time.
Thanks for the advice.
Ron
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Exporting Data From MySQL Using PHP
On Tuesday 10 August 2004 11:34, Peter Ellis wrote: It has been my experience that you sometimes need to do backquotes (`) in MySQL queries when cutting/pasting into PHP -- it's the same key as the tilde (~) on my keyboard. Try: $result = mysql_query(SELECT e_mail FROM subscriptionsdatabase WHERE `discipleship_mailing_list_e_mail_subscription` LIKE 'on'); The backticks (`) are only needed if: (a) your column names uses (MySQL) reserved words (eg names of the builtin functions etc) (b) you have spaces in your column names Both (a) and (b) are rather bad and shouldn't really be used. Just use 'friendly' column names and backticks will not be necessary. That should work. I've always used the equal sign instead of LIKE, but that seems to be a matter of preference. It is not a matter of preference, = and LIKE have different purposes. Refer to manual and/or some SQL tutorial for details. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* Why won't you let me kiss you goodnight? Is it something I said? -- Tom Ryan */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Exporting Data From MySQL Using PHP
Ron Piggott wrote: I was able to get the command: SELECT e_mail FROM subscriptionsdatabase WHERE discipleship_mailing_list_e_mail_subscription LIKE 'on' to work in the mySQL command prompt but when I put it into a .PHP file I get a parse error. I get that parse error by simply coping and pasting the command into the PHP file. The reason that this didn't work is that you have to tell PHP first to create a link with your database, then execute a query and get back the results by issuing a few specific commands. Robby's example shows you how, but it's more thorougly explained in the php documentation. Have a look at http://www.php.net/manual/en/ref.mysql.php and its subpages. It has examples and explanations you'll find very useful in setting up PHP to work with a MySQL database. Regards, Guus der Kinderen -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: Exporting Data From MySQL Using PHP
It seems to me that, at least in terms of consistency, using backticks isn't a bad idea. After all, if you look at any current version of phpMyAdmin that generates PHP selection code, it uses backticks constantly. I take your point, but it seems to me like it's a good habit no matter what. Perhaps that's more a matter of semantics than PHP programming itself :) As for the difference between = and LIKE, you are correct -- I just don't typically use LIKE for any of my database work, so I forget there's a difference every once in a while :) Thank you for the reminder. -- Peter Ellis - [EMAIL PROTECTED] Web Design and Development Consultant naturalaxis | http://www.naturalaxis.com/ On Tue, 2004-08-10 at 12:33 +0800, Jason Wong wrote: On Tuesday 10 August 2004 11:34, Peter Ellis wrote: It has been my experience that you sometimes need to do backquotes (`) in MySQL queries when cutting/pasting into PHP -- it's the same key as the tilde (~) on my keyboard. Try: $result = mysql_query(SELECT e_mail FROM subscriptionsdatabase WHERE `discipleship_mailing_list_e_mail_subscription` LIKE 'on'); The backticks (`) are only needed if: (a) your column names uses (MySQL) reserved words (eg names of the builtin functions etc) (b) you have spaces in your column names Both (a) and (b) are rather bad and shouldn't really be used. Just use 'friendly' column names and backticks will not be necessary. That should work. I've always used the equal sign instead of LIKE, but that seems to be a matter of preference. It is not a matter of preference, = and LIKE have different purposes. Refer to manual and/or some SQL tutorial for details. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* Why won't you let me kiss you goodnight? Is it something I said? -- Tom Ryan */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Posting Data to MySQL
Tom Chubb [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] I am trying to design a form that posts the data to a DB, but being new to PHP/MySQL, it takes me ages to hand write the code and I'm sure there must be an easier way to do it? My form has 100 fields using 20 rows and 5 columns and it's taking me ages to write the code for it. Can anyone advise how they code large forms like that and if you know of any software to make it easier? I've searched Google for hours and checked loads of Database scripts on sites like Hotscripts, but can't find anything. Hi Tom, that's a lot of information to display on one page. Can you split it up into several forms? Of course this would not mean fewer work but maybe make it a bit mor concise. Regards, Torsten Roehr -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: Posting Data to MySQL
Doesn't matter how many fields are on a form; they are returned as an array of strings in the request object. Just iterate through the array and update the corresponding database fields. Mark -Original Message- From: Torsten Roehr [mailto:[EMAIL PROTECTED] Sent: Thursday, June 24, 2004 10:02 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Re: Posting Data to MySQL Tom Chubb [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] I am trying to design a form that posts the data to a DB, but being new to PHP/MySQL, it takes me ages to hand write the code and I'm sure there must be an easier way to do it? My form has 100 fields using 20 rows and 5 columns and it's taking me ages to write the code for it. Can anyone advise how they code large forms like that and if you know of any software to make it easier? I've searched Google for hours and checked loads of Database scripts on sites like Hotscripts, but can't find anything. Hi Tom, that's a lot of information to display on one page. Can you split it up into several forms? Of course this would not mean fewer work but maybe make it a bit mor concise. Regards, Torsten Roehr -- 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] Re: Posting Data to MySQL
Mark A Galbreath [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Doesn't matter how many fields are on a form; they are returned as an array of strings in the request object. Just iterate through the array and update the corresponding database fields. Mark, what I was trying to say was that having 100 form elements on one page is quite a lot of information to comprehend for the user. And if each form element has a different name you cannot just handle the data as an array. You have to address each value individually. Regards, Torsten -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Retrieve data from a table, edit/add it and enter it in a new table
I see one bug in your code , that is you never rewind the pointer of
mysql_fetch_array($result), so at the end of the first cycle ...while
($r=mysq...)... the pointer is at the end of the query resource. You should
use mysql_data_seek($result,0) to rewind before doing another while cycle.
Hope it can help
Bye
Francesco Basile (PHP_newbee)
Justin [EMAIL PROTECTED] ha scritto nel messaggio
news:[EMAIL PROTECTED]
Hi,
I am trying to do the following:
Retrieve some information from a table, edit it by appending some further
information to it (a few more fields) and then enter the new data record
into a new table, and delete the old data in the original table. Sounds
confusing I know.
The code is below (I apologise for its poor style etc as I am very new to
php), but when I click on the 'Enter Information' button, nothing happens.
The existing data is retrieved without any problems and I can select it
using the radio button. But when I try and add data to the new fields
('option_close_price' and 'notes'), nothing happens.
Any help appreciated.
form action=? echo $PHP_SELF ? method=post
?
mysql_pconnect(localhost,root,password);
mysql_select_db(options);
if(!$cmd)
{
$result = mysql_query(select * from open_trades);
while($r=mysql_fetch_array($result))
{
$open_date=$r[open_date];
$share=$r[share];
$code=$r[code];
$short_long_trade=$r[short_long_trade];
$id=$r[id];
$expiry=$r[expiry];
$exercise=$r[excercise];
$option_price=$r[option_price];
$no_purchased=$r[no_purchased];
$no_sold=$r[no_sold];
$income_in=$r[income_in];
$income_out=$r[income_out];
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print td/tdtdOpen Date/tdtdShare/tdtdCode/tdtdShort
orbr Long Trade/tdtdExpiry/tdtdExcercise/tdtdOption
Price/tdtdNumberbr Purchased/tdtdNumber Sold/tdtdIncome
In/tdtdIncome Out/tdtd
/tr;
while ($row = mysql_fetch_array($result))
{
print trtd;
print INPUT TYPE='RADIO' NAME='id' VALUE='echo $id';
print /tdtd;
print $row[open_date];
print /tdtd;
print $row[share];
print /tdtd;
print $row[code];
print /tdtd;
print $row[short_long_trade];
print /tdtd;
print $row[expiry];
print /tdtd;
print $row[excercise];
print /tdtd;
print $row[option_price];
print /tdtd;
print $row[no_purchased];
print /tdtd;
print $row[no_sold];
print /tdtd;
print $row[income_in];
print /tdtd;
print $row[income_out];
print /td/tr\n;
}
print /table\n;
?
? }?
input type=submit name=cmd value=Close/form
? }
?
?
if($cmd==Close)
{
if (!$submit)
{
$result = mysql_query(select * from open_trades);
while($myrow=mysql_fetch_array($result))
?
input type=hidden name=id value=?php echo $myrow[id] ?
Option Close PriceINPUT TYPE=TEXT NAME=option_close_price VALUE=?PHP
echo $myrow[option_close_price]? SIZE=7 br
Notes:INPUT TYPE=TEXT NAME=notes VALUE=?php echo $myrow[notes] ?
SIZE=60br
input type=hidden name=cmd value=edit
input type=Submit name=submit value=Enter information
/form
? } ?
?
if($submit)
{
$query = INSERT INTO closed_trades SET code='$code',
option_price='$option_price', option_close_price='$option_close_price',
no_sold='$no_sold', open_date='$open_date',
short_long_trade='$short_long_trade, share='$share', expiry='$expiry',
excercise='$excercise', no_purchased='$no_purchased',
income_in='$income_in', income_out='$income_out', notes='$notes', id=$id';
$result = mysql_query($ql);
$query = DELETE FROM open_trades WHERE id=$id;
$result = mysql_query($sql);
echo Thank you! Information updated.;
}
}
?
/td
/table
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[PHP-DB] Re: Getting data from mysql db for html form
works like a bomb !!!
thanx man !
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Gamco - Gawie Marais [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi,
Could someone help me out here...
I have a mysql database with a table that conatains names. I am setting
up
a
html form and I would like to create a multi-listing with those names so
that you can choose one and then submit the form.
// open db connection...
$result = mysql_query(SELECT name FROM table);
echo 'select name=names';
while ($row = mysql_fetch_assoc($result)) {
echo 'option' . $row['name'] . '/option';
}
echo '/select';
Haven't tested it. Change 'name' to the name of your table column.
Regards,
Torsten Roehr
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[PHP-DB] Re: Getting data from mysql db for html form
Gamco - Gawie Marais [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi,
Could someone help me out here...
I have a mysql database with a table that conatains names. I am setting up
a
html form and I would like to create a multi-listing with those names so
that you can choose one and then submit the form.
// open db connection...
$result = mysql_query(SELECT name FROM table);
echo 'select name=names';
while ($row = mysql_fetch_assoc($result)) {
echo 'option' . $row['name'] . '/option';
}
echo '/select';
Haven't tested it. Change 'name' to the name of your table column.
Regards,
Torsten Roehr
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[PHP-DB] Re: Dynamic Data
Christopher J. Crane wrote:
What is the best way to produce a report listing the fieldname and then the
data in that field. It is a report containing only one row from a table, but
I don't want to hard code the fields since they change often.
I can get the field names dynamically like this:
$fields = mysql_list_fields(Network, Subnets);
$num_columns = mysql_num_fields($fields);
for($ i = 0; $i $num_columns; $i++) { echo mysql_field_name($fields,
$i); }
now is where I get confusedwhat I would like to happen is do a query on
a single row and put the results into mysql_fetch_assoc and during the prior
loop put the column name into the loop of the data. Something like:
$fields = mysql_list_fields(Network, Subnets);
$num_columns = mysql_num_fields($fields);
$result = mysql_query(SELECT * FROM table1 WHERE ID = '$ID';
$field = mysql_fetch_assoc($result)
for($ i = 0; $i $num_columns; $i++) {
echo b . mysql_field_name($fields, $i) . : /b .
$field[mysql_field_name($fields, $i)] . br\n;
All you really need here is:
$sth = mysql_query(SELECT * FROM table1 WHERE ID = '$ID');
$rec = mysql_fetch_assoc($sth);
foreach($rec as $field = $value) {
echo 'b'.$field.': /b'.$value.'br/';
}
--
paperCrane Justin Patrin
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[PHP-DB] Re: inserting data into database!
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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[PHP-DB] Re: inserting data into database!
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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Re: [PHP-DB] Re: inserting data into database!
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
If this is the actual code, you are missing a closing brace at the end of
the block.
dave
[PHP-DB] Re: inserting data into database!
Now that I closed the bracket
it tells me that $row is not defined.
I have plagiarized from a couple of different places to come up with my code
so that is the reason that it may not work that well. Here is my code right
now
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
}
?
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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RE: [PHP-DB] Re: inserting data into database!
Have you declared a new instance of the db class anywhere on your page?
-Original Message-
From: Ronald Allen [mailto:[EMAIL PROTECTED]
Sent: Friday, May 14, 2004 4:13 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: inserting data into database!
Now that I closed the bracket
it tells me that $row is not defined.
I have plagiarized from a couple of different places to come
up with my code
so that is the reason that it may not work that well. Here
is my code right
now
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
}
?
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an
auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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RE: [PHP-DB] Re: inserting data into database!
...or whatever class contains $row- functionality?
-Original Message-
From: Ronald Allen [mailto:[EMAIL PROTECTED]
Sent: Friday, May 14, 2004 4:13 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: inserting data into database!
Now that I closed the bracket
it tells me that $row is not defined.
I have plagiarized from a couple of different places to come
up with my code
so that is the reason that it may not work that well. Here
is my code right
now
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
}
?
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an
auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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Re: [PHP-DB] Re: inserting data into database!
Go to PHP.net, books @ bookstore.
Seems like you are trying to insert data from an array into a MySQL table, I
would visit http://dev.mysql.com/doc/mysql/en/index.html site too.
Good luck.
===
Ronald The Newbie Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
I have no idea what a class is . I really am a newbie. After you guys
help
me with this what is a good place to start with learning PHP?
Richard Hutchins [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
...or whatever class contains $row- functionality?
-Original Message-
From: Ronald Allen [mailto:[EMAIL PROTECTED]
Sent: Friday, May 14, 2004 4:13 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: inserting data into database!
Now that I closed the bracket
it tells me that $row is not defined.
I have plagiarized from a couple of different places to come
up with my code
so that is the reason that it may not work that well. Here
is my code right
now
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
}
?
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Here is my code now, but I am getting a parsing error
Parse error: parse error, unexpected $end
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
?
Torsten Roehr [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Ronald Allen [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Two questions?
First why am I getting undefined variable tmp , yet it works
Second how would I insert data into this table?
?
$db_table = 'log';
$conn = mysql_connect($db_host,$db_user,$db_pass);
if ($conn == true) {
mysql_select_db($db_name,$conn);
$result = mysql_query(SELECT * from $db_table,$conn);
while($row = mysql_fetch_object($result)) {
$tmp .= ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and_Time br\n;
$tmp .= Event_Type : $row-Event_Type br\n;
$tmp .= Description : $row-Description br\n;
$tmp .= Initials : $row-Initials brhr\n;
}
} else {
echo 'could not connect to database : '. mysql_error();
}
print $tmp;
?
You have to initialize $temp before using it with '.='
$temp = '';
$temp .= 'my string';
Or just set the first line with =
$tmp = ID : $row-ID br\n;
$tmp .= Base : $row-Base br\n;
$tmp .= Date_and_Time : $row-Date_and
Your insert should look like this (if ID is an
auto-increment field):
$insert = INSERT INTO $db_table SET ;
$insert .= Base = ' . $row-Base . ', ;
$insert .= Date_and_Time = ' . $row-Date_and_Time . ', ;
$insert .= Event_Type = ' . $row-Event_Type . ', ;
$insert .= Description = ' . $row-Description . ', ;
$insert .= Initials = ' . $row-Initials . ';
mysql_query($insert, $conn);
Hope this helps,
Regards, Torsten
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[PHP-DB] Re: finding data
the way I would do this is with count()
$row = mysql_fetch_assoc(SELECT count(*) as c FROM bivalues WHERE
myvalue = '$zchar')
if ($row['c'] 0)
{
echo 'result 2 - found it'
}else{
echo 'NOT found'
}
hope it helps
Pete
Douglas D Hull wrote:
I am trying to find data, or if not found set a variable to Err. myvalue being a variable in my database and $zchar being my value to find.
The following result is never 1 even if no values were found, so the result is always
2. If the finding value does exist in my database this does find the correct value:
if (!$result = mysql_query(SELECT * FROM bivalues WHERE myvalue = '$zchar'))
{
echo 'result 1 - not found';
$myresut = Err;
}
else
{
{echo 'result 2 - found it'
}
The following result is never 2 even if the value actually exist in my database, so
the result is always 1:
if (!$result = mysql_query(SELECT * FROM bivalues WHERE myvalue = $zchar))
{
echo 'result 1 - not found';
$myresult = Err;
}
else
{
{echo 'result 2 - found it'
}
Thanks for any help,
Doug
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Re: [PHP-DB] Re: finding data
The way I would do it:
$result = mysql_query(select * from bivalues WHERE myvalue - '$zchar');
if (mysql_num_rows($result) {
echo Result 2 - found it;
} else {
echo Not found;
}
This leaves you with the option to start processing the data as soon as
error checking is over. Using Count, you'll have to re-retrieve the
data from the database, which takes twice as long time as doing it only
once.. (or thereabout).
The problem you had is that no matter if the query presents results or
not, the result variable is only set to 0 (thus the !$result syntax
gives results) when there's been an error in the mysql query. Simply an
empty table will not return an error!
Mike
On Apr 15, 2004, at 16:54, pete M wrote:
the way I would do this is with count()
$row = mysql_fetch_assoc(SELECT count(*) as c FROM bivalues WHERE
myvalue = '$zchar')
if ($row['c'] 0)
{
echo 'result 2 - found it'
}else{
echo 'NOT found'
}
hope it helps
Pete
Douglas D Hull wrote:
I am trying to find data, or if not found set a variable to Err.
myvalue being a variable in my database and $zchar being my value to
find.
The following result is never 1 even if no values were found, so the
result is always 2. If the finding value does exist in my database
this does find the correct value:
if (!$result = mysql_query(SELECT * FROM bivalues WHERE myvalue =
'$zchar'))
{
echo 'result 1 - not found';
$myresut = Err;
}
else
{
{echo 'result 2 - found it'
}
The following result is never 2 even if the value actually exist in
my database, so the result is always 1:
if (!$result = mysql_query(SELECT * FROM bivalues WHERE myvalue =
$zchar))
{
echo 'result 1 - not found';
$myresult = Err;
}
else
{
{echo 'result 2 - found it'
}
Thanks for any help,
Doug
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[PHP-DB] Re: insert data in mysql table
From what I can see, the order that you created the table with and the
order of the fields in the query do not match up. Changing the query to:
$add_all = INSERT INTO $table
values('','$name','$day','$question','$email','');
Should make things run just fine.
Cheers,
Andy
Charalambos Nicolaou wrote:
Hi everyone ,
I have created this Mysql table
CREATE TABLE questions (ID INT NOT NULL AUTO_INCREMENT, name
VARCHAR(25), day DATE, question TEXT, email VARCHAR(30), PRIMARY KEY(ID));
And I am trying to insert data in it using the following
body
?
if(isset($_GET['commented']))
{
echo(Your comment has been posted.);
$host = ;
$user = **;
$pass = *;
$db = **;
$table = questions;
$name = $_GET['form_uname'];
$question = $_GET['form_quest'];
$day = $_GET['form_day'];
$email = $_GET['from_email'];
mysql_connect($host,$user,$pass) or die(mysql_error());
mysql_select_db($db) or die(mysql_error());
$add_all = INSERT INTO $table
values('$name','$question','$day','$email','');
mysql_query($add_all) or die(mysql_error());
}
else
{
?
form method=get action=? echo $PHP_SELF ; ?
Name : input type=text name=form_uname
pQuestion :textarea rows=3 name=form_quest cols=36/textarea/p
pbr
Date: input type=text name=form_day
/p
pEmail: input type=text name=form_emailbrbr
input type= hidden name= commented value= set
input type=submit value=Submit
/p
/form
?php
}
?
/body
Finally has a result to get the correct data only for the DATE and the
data from the Question FORM goes to the name field. I do not know where
the problem is. Help me please because I am going crazy,
Thanks in advance
Charalambos Nicolaou
_
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RE: [PHP-DB] Re: exporting data to excel
Good stuff to know! Thx, Geir! Mark -Original Message- From: Geir Pedersen - Activio AS [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 24, 2004 9:05 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Re: exporting data to excel Matthew, I am looking for the easiest way to export data to an excel file. Is the eaiest way to use PHP's file handling functions? I assume you want to delivery the execl file to a web user. Then there is no need to write the data to disk, you can generate the whole thing on the fly while you deliver it to the user. Here is a basic description on how: 1) Create a PHP document to generate the execl file and make sure the script produces HTTP headers telling the user agent that it is receiving an excel document. Start the script with: header (Content-type: application/vnd.ms-excel); header (Content-Disposition: attachment ); The content disposition header says that the execl file is to be stored on disk. 2) Generate the spreadsheet data as a list of newline separated rows with tab separated fields: echo field1\tfield2\t...\n; 3) Set up a link to the new PHP document. When a user clicks on that link, he will be prompted by his user agent on where to save the excel file. --- Geir Pedersen http://www.activio.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] Re: exporting data to excel
Hello all
I am also needing to do this - I got the xcel ss to generate but it wont tab to the
next cell in xcel...
Here's what I've tried so far:
?php
while ($row = mysql_fetch_assoc($res)) {
echo $row[first] . \t . $row[last] . \n;
// echo $row[first] . , . $row[last] . \n;
// echo $row[first],$row[last]\n;
// echo $row[first]\t$row[last]\t\n;
// echo $row[first] . , . $row[last];
}
?
***
But I get all the fields in one cell on each of these tries...
Thanks
Geir Pedersen - Activio AS [EMAIL PROTECTED] 03/24/04 08:04PM
Matthew,
I am looking for the easiest way to export data to an excel file. Is
the eaiest way to use PHP's file handling functions?
I assume you want to delivery the execl file to a web user.
Then there is no need to write the data to disk, you can
generate the whole thing on the fly while you deliver it to the
user. Here is a basic description on how:
1) Create a PHP document to generate the execl file and make
sure the script produces HTTP headers telling the user agent
that it is receiving an excel document. Start the script
with:
header (Content-type: application/vnd.ms-excel);
header (Content-Disposition: attachment );
The content disposition header says that the execl file is
to be stored on disk.
2) Generate the spreadsheet data as a list of newline separated
rows with tab separated fields:
echo field1\tfield2\t...\n;
3) Set up a link to the new PHP document. When a user clicks on
that link, he will be prompted by his user agent on where to
save the excel file.
---
Geir Pedersen
http://www.activio.com/
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[PHP-DB] Re: exporting data to excel
Here's what I ended up doing that works:
(do all the db connection stuff - get your result, etc..)
table width=100% border=1
tr
td div align=centerstrongName/strong/div/td
td div align=centerstrongCompany /strong/div/td
td div align=centerstrongAddress /strong/div/td
td div align=centerstrongE-mail/strong/div/td
td div align=centerstrongTelephone/strong/div/td
/tr
?php
while ($row = mysql_fetch_assoc($res)) {
?
tr
td?php echo $row[first] . $row[last]; ?/td
td?php echo $row[company] . $row[title]; ?/td
td?php echo $row[address1] . , . $row[city] . , . $row[state] . , .
$row[zip]; ?/td
td?php echo $row[email]; ?/td
td?php echo $row[phone]; ?/td
/tr
?php
}
?
/table
**
use the header() info in this post at the top of your page, create another page and
link to this script. You can either view the excel or save to disk...
Thanks
Mignon
Geir Pedersen - Activio AS [EMAIL PROTECTED] 03/24/04 08:04PM
Matthew,
I am looking for the easiest way to export data to an excel file. Is
the eaiest way to use PHP's file handling functions?
I assume you want to delivery the execl file to a web user.
Then there is no need to write the data to disk, you can
generate the whole thing on the fly while you deliver it to the
user. Here is a basic description on how:
1) Create a PHP document to generate the execl file and make
sure the script produces HTTP headers telling the user agent
that it is receiving an excel document. Start the script
with:
header (Content-type: application/vnd.ms-excel);
header (Content-Disposition: attachment );
The content disposition header says that the execl file is
to be stored on disk.
2) Generate the spreadsheet data as a list of newline separated
rows with tab separated fields:
echo field1\tfield2\t...\n;
3) Set up a link to the new PHP document. When a user clicks on
that link, he will be prompted by his user agent on where to
save the excel file.
---
Geir Pedersen
http://www.activio.com/
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[PHP-DB] Re: exporting data to excel
Matthew, I am looking for the easiest way to export data to an excel file. Is the eaiest way to use PHP's file handling functions? I assume you want to delivery the execl file to a web user. Then there is no need to write the data to disk, you can generate the whole thing on the fly while you deliver it to the user. Here is a basic description on how: 1) Create a PHP document to generate the execl file and make sure the script produces HTTP headers telling the user agent that it is receiving an excel document. Start the script with: header (Content-type: application/vnd.ms-excel); header (Content-Disposition: attachment ); The content disposition header says that the execl file is to be stored on disk. 2) Generate the spreadsheet data as a list of newline separated rows with tab separated fields: echo field1\tfield2\t...\n; 3) Set up a link to the new PHP document. When a user clicks on that link, he will be prompted by his user agent on where to save the excel file. --- Geir Pedersen http://www.activio.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Inserting Data in Multiple Tables - MySQL
Craig Hoffman wrote: Is it possible to write one INSERT statement to populate multiple tables? __ Craig Hoffman - eClimb Media v: (847) 644 - 8914 f: (847) 866 - 1946 e: [EMAIL PROTECTED] w: www.eclimb.net _ Nope. One record, one insert. Well, actually, some DBs support multiple inserts with one query (such as mysql 4) but not into multiple tables. -- paperCrane Justin Patrin -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Inserting Data in Multiple Tables.. again
Craig Hoffman wrote: Here is the problem I am experiencing: I have a form that collects data and I would like it to send it to three tables in a MySQL DB. One table is called TRAININGLOG , other is called CORE and the third is USERS. The USER table is the main table that contains the primary key. My question is, what is the correct way for handling something like this? Do I need to write three separate queries (INSERT) statements? I am using MySQL 4.0.17 if that helps? Thanks in advance, Craig __ Craig Hoffman - eClimb Media v: (847) 644 - 8914 f: (847) 866 - 1946 e: [EMAIL PROTECTED] w: www.eclimb.net _ You need three seperate queries. -- paperCrane Justin Patrin -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Storing data in a file
Todd Cary wrote: I have a client that does not want to use a DBMS for storing the email addresses the surfers input. Is there a class available that has a search function as well as an append? I would like to have several fields separated by a tab or : . Todd I have heard that SQLite is pretty nice in this regard. It implements a basic database system but without having to run a seperate database process. (Honestly, running mysql isn't that hard, but...) -- paperCrane Justin Patrin -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] RE: show data from 2 tables
Try rewriting your query. It looks to me that you probably have an
error in your SQL statement and it only seems fine because you appended
the @ to the beginning of the mysql_query() function.
I would suggest using a join statement.
If(!$result = mysql_query(SELECT orders.orderid, orders.amount,
orders.date, customer.name FROM orders RIGHT JOIN customers on
orders.customerid = customers.customerid WHERE customers.name = 'Don
Hansen')) {
echo(mysql_error()); //I would take this out after you are done
debuging
} else {
while ($row = mysql_fetch_array($result)) {
$zorderid = $row['orders.orderid'];
$zamt = $row['orders.amount'];
echo(str_pad($zorderid,4,' ').str_pad($zamt,6, ' ').'br');
}
}
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 28, 2003 9:20 AM
To: [EMAIL PROTECTED]
Subject: show data from 2 tables
I am attempting to pull data from 2 tables. My SELECT statement seems
ok. But the next line: while ... etc. is giving me an error of
Supplied argument is not a valid MySQL result resource.. But this is
the way I usually show my data and it works fine when only 1 table is
involved. Is there a different way of showing your data when pulling
data from more than one table?
$result = @mysql_query('SELECT orders.orderid, orders.amount,
orders.date
from customers, orders WHERE customers.name = Don Hansen and
customers.customerid = orders.customerid');
while ( $row = mysql_fetch_array($result) )
{
$zorderid = $row['orders.orderid'];
$zamt = $row['orders.amount'];
echo (str_pad($zorderid,4,' ') . str_pad($zamt,6, ' ') .
'br');
}
Thanks,
Doug
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[PHP-DB] Re: copying data
If its a 'one off' kind of transfer then i'd recommend that you consider
doing what I did - export the data from Access to text file(s) and then
import from the text file(s) into MySQL (phpMyAdmin is v handy for this lind
of task) {Of course if you want a more automated solution then this is not
the way to go)
David Eisenhart
If I have a microsoft access file .mdb on my server, can I use a php page
to move the data from the file to a mysql db?
Thanks,
Eddie
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[PHP-DB] Re: copying data
If its a 'one off' kind of transfer then i'd recommend that you consider
doing what I did - export the data from Access to text file(s) and then
import from the text file(s) into MySQL (phpMyAdmin is v handy for this lind
of task) {Of course if you want a more automated solution then this is not
the way to go)
David Eisenhart
If I have a microsoft access file .mdb on my server, can I use a php page
to move the data from the file to a mysql db?
Thanks,
Eddie
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[PHP-DB] RE: copying data
There is a utility called MyAcess it is like MyODBC and I have used it to do
this exact thing. It works very well. I believe you can find it on
Freshmeat.
Thomas G. Knight
[EMAIL PROTECTED]
http://www.slaponline.com
-Original Message-
From: David Eisenhart [mailto:[EMAIL PROTECTED]]
Sent: Sunday, December 15, 2002 4:51 AM
To: [EMAIL PROTECTED]
Subject: Re: copying data
If its a 'one off' kind of transfer then i'd recommend that you consider
doing what I did - export the data from Access to text file(s) and then
import from the text file(s) into MySQL (phpMyAdmin is v handy for this lind
of task) {Of course if you want a more automated solution then this is not
the way to go)
David Eisenhart
If I have a microsoft access file .mdb on my server, can I use a php page
to move the data from the file to a mysql db?
Thanks,
Eddie
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[PHP-DB] Re: upload data to MySql
Thanx, I found out I forgot to pass it to my db.
Jacco
--
http://seabird.jmtech.ca
Attitude is Everything!
But Remember, Attitudes are Contagious!
Is Yours worth Catching
David Robley [EMAIL PROTECTED] wrote in message
news:MPG.183f2f6ec46a84169897c8;news.php.net...
In article [EMAIL PROTECTED], [EMAIL PROTECTED]
says...
Hi everyone,
I don't get my upload to work properly. It's a tutorial of the web, but
it
doesn't function (unless I made a mistake). Please help me...
I have a form passing on 5 fields (text for testing), name: year, make,
model, price, picture and submit to PHP_SELF
?php
if ($submit) {
$db = mysql_connect(localhost,myname,mypassword);
mysql_select_db($test,$db);
//DB Test for testing...
$sql = INSERT INTO test (year,make,model,price,picture) VALUES
('$year,$make,$model,$price,$picture');
This line looks like your first problem. Each value being inserted which
is a text type should be separately surrounded by single quotes. So:
('$year','$make','$model','$price','$picture');
As a debugging aid, use
echo mysql_error()
after you pass the query to mysql with mysql_query() I don't see where you
actually do that in this code snippet??
//I created these columns in the table test
echo year: $yearbr\n;
echo make: $makebr\n;
echo model: $modelbr\n;
echo price: $pricebr\n;
echo picture: $picturebr\n;
}
?
I know I need globals on for this.
Thanx for the help,
--
David Robley
Temporary Kiwi!
Quod subigo farinam
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[PHP-DB] Re:[PHP-DB] Re: Print data
I'm interested in examples with codes thanks in advance.. From: Dan Koken [EMAIL PROTECTED] Date: 2002/10/25 Fri AM 06:12:55 GMT+03:00 To: [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: [PHP-DB] Re: Print data We faced this problem, and decided if we couldn't get a good solution to eliminate the printer. Turns out the users liked it better, and we eliminated a lot of paper. - First we wrote a report writer. - All reports go to the database. - We wrote a report file system. - The report output title shows up in the user report file system in the IN BASKET. - The user can click the file and it displays the first 5 pages to the screen. They can change the start page and number of pages. - The user can file it into other files like WEEKLY REPORTS. - They can send the report to others and it shows up in their IN BASKET. - If they delete the report it sits in the WASTE BASKET file for 10 days. This gives them the ability to file it back somewhere if it was deleted by mistake. This is fast because the reports are run on the server. Only one copy of the report is kept in the database, and the user report file system points to it. Anyway this was our solution and the users love it since they no longer have to file paper, but file reports electronically. They can carbon copy anyone. They can look at any section of the report any time. They can still print the report if they want by using the print button. HTH... Have a great day... Dan José Moreira wrote: hello, at my company we have several network printers and i was was wondering if it's possible to print directly to them using PHP, instead of showing on the screen or both ... to gain more control on what is printed, because of the nasty browsers header and footer ... whom i now how to remove ... i know that the printer IP is 192.192.1.100 and the port is 9005 ... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Free Email Account at www.flash.ro -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Print data
We faced this problem, and decided if we couldn't get a good solution to eliminate the printer. Turns out the users liked it better, and we eliminated a lot of paper. - First we wrote a report writer. - All reports go to the database. - We wrote a report file system. - The report output title shows up in the user report file system in the IN BASKET. - The user can click the file and it displays the first 5 pages to the screen. They can change the start page and number of pages. - The user can file it into other files like WEEKLY REPORTS. - They can send the report to others and it shows up in their IN BASKET. - If they delete the report it sits in the WASTE BASKET file for 10 days. This gives them the ability to file it back somewhere if it was deleted by mistake. This is fast because the reports are run on the server. Only one copy of the report is kept in the database, and the user report file system points to it. Anyway this was our solution and the users love it since they no longer have to file paper, but file reports electronically. They can carbon copy anyone. They can look at any section of the report any time. They can still print the report if they want by using the print button. HTH... Have a great day... Dan José Moreira wrote: hello, at my company we have several network printers and i was was wondering if it's possible to print directly to them using PHP, instead of showing on the screen or both ... to gain more control on what is printed, because of the nasty browsers header and footer ... whom i now how to remove ... i know that the printer IP is 192.192.1.100 and the port is 9005 ... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Print data
an alternative I'm looking into is simply writing/formatting the data into a text file, then using the good old 'net print' dos command to print the file... José Moreira [EMAIL PROTECTED] wrote in message news:1035381924.1052.8.camel;inf4.pt... hello, at my company we have several network printers and i was was wondering if it's possible to print directly to them using PHP, instead of showing on the screen or both ... to gain more control on what is printed, because of the nasty browsers header and footer ... whom i now how to remove ... i know that the printer IP is 192.192.1.100 and the port is 9005 ... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Print data
José Moreira wrote: hello, at my company we have several network printers and i was was wondering if it's possible to print directly to them using PHP, instead of showing on the screen or both ... to gain more control on what is printed, because of the nasty browsers header and footer ... whom i now how to remove ... i know that the printer IP is 192.192.1.100 and the port is 9005 ... If you're willing to do a bit of work, you can do a workable solution. I'm assuming that PHP is running on a Windows server. You can use the Windows printer functions if the printer was connected directly. If not, you can still get it to work. The Windows Scripting Host resources give you a few ways. A kludgy way would be to configure IE on the printer machine (since it's almost surely installed) to not show the header and footers. You can use the WSH functions to open a non-visible IE instance and issue the print command from there. WSH also has functions in it's COM objects for mapping network printers, etc so you could customize your own server's IE config and print to a mapped printer. Lots of ways to skin this cat, but they're all messy. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Accessing data from next row? (mysql)
Create an array and go through the array as needed.
Here is some code from one of my db functions:
$DB_RESULT = mysql_query($sql) or die(Error executing query: .
mysql_error());
// create an empty array to fill with data
$arrData = array();
$rowCount = 0;
while ($r = mysql_fetch_array($DB_RESULT, MYSQL_ASSOC)){
foreach ($r as $key = $value){
$arrData[$rowCount][$key] = $value;
}
$rowCount++;
}
then simply instead of your code:
while($array = mysql_fetch_array($result)){
use this:
for ($i=0;$icount($arrData);$i++){
$arrThisRow = $arrData[$i];
$arrNextRow = $arrData[$i+1];
}
obviously, if you are trying to access a variable which index does not
exist, you will need to implement some sort of error checking, etc
adam
Using mysql, how do I access the data of the next row using code
something like this:
$result = mysql_query(select column from table where
whatever='whatever');
while($array = mysql_fetch_array($result)){
//Whatever
}
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[PHP-DB] Re: Accessing data from next row? (mysql)
Also, if you're not sure how to get the values from $arrThisRow /
$arrNextRow:
echo $arrThisRow['columnName'];
or $arrData[$i]['columnName'] if you want to access it manually without
the other two arrays
adam
On Monday, October 14, 2002, at 06:05 PM, Adam Royle wrote:
Create an array and go through the array as needed.
Here is some code from one of my db functions:
$DB_RESULT = mysql_query($sql) or die(Error executing query: .
mysql_error());
// create an empty array to fill with data
$arrData = array();
$rowCount = 0;
while ($r = mysql_fetch_array($DB_RESULT, MYSQL_ASSOC)){
foreach ($r as $key = $value){
$arrData[$rowCount][$key] = $value;
}
$rowCount++;
}
then simply instead of your code:
while($array = mysql_fetch_array($result)){
use this:
for ($i=0;$icount($arrData);$i++){
$arrThisRow = $arrData[$i];
$arrNextRow = $arrData[$i+1];
}
obviously, if you are trying to access a variable which index does not
exist, you will need to implement some sort of error checking, etc
adam
Using mysql, how do I access the data of the next row using code
something like this:
$result = mysql_query(select column from table where
whatever='whatever');
while($array = mysql_fetch_array($result)){
//Whatever
}
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[PHP-DB] Re: odbc data sort problem (i think) :)
i got the problem. the odbc driver fills unuses spaces in a row with blanks. goddamn *g* cu Lukas Boldrino [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... hey guys ! I´m trying to use a if syntax to compare data and preselect radiobuttons. I tried it like that: if ($checked == Patch) $checked6 = bla; With numbers it works great but with char my $checked6 variable is empty ! I know php is case sensitive and I´m sure that problem is not the case sensitive so please help, im using a MS SQL Server 2000 (enterprise edition) and the odbc driver! thanks guys! - [EMAIL PROTECTED] - -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: odbc data sort problem (i think) :)
Lukas, Use varchar instead of char to avoid this issue - that is the behavior you should expect regardless of the driver or ODBC use. Best regards, Andrew Hill OpenLink Software On Thursday, August 29, 2002, at 05:07 AM, Lukas Boldrino wrote: i got the problem. the odbc driver fills unuses spaces in a row with blanks. goddamn *g* cu Lukas Boldrino [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... hey guys ! I¥m trying to use a if syntax to compare data and preselect radiobuttons. I tried it like that: if ($checked == Patch) $checked6 = bla; With numbers it works great but with char my $checked6 variable is empty ! I know php is case sensitive and I¥m sure that problem is not the case sensitive so please help, im using a MS SQL Server 2000 (enterprise edition) and the odbc driver! thanks guys! - [EMAIL PROTECTED] - -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: change data before its sent to db?
You will laugh. When you quote month, you don't sort date by month field. You sort by month - text constant being always equal. In fact you son't sort it at all! The sort code --- if ($orderby == 'month_num'): $sql = select * from releases order by 'month'; elseif ($orderby == 'month_num2'): $sql = select * from releases order by 'month' desc; Try this: $sql = select * from releases order by month; -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: extract data from database into an array
Chip Wiegand [EMAIL PROTECTED] wrote
I have a database, all the data is numbers. I want to make a query that
will extract the data and then make it available in an array, so the
array is populated 'real-time'. I could just enter the number into an
array manually, but I want to automate the job. I don't know what to
start looking for, the 3 books I have either don't talk about this or I
just am missing it.
Could someone point me in the right direction?
This class makes it possible to return an array in two ways very easy.
Just try this
$db = new Ms_db;
$db-connect(true); // open a persistant connection
$vars1 = $db-query(SELECT * FROM table);
$vars2 = $db-query(SELECT * FROM table,off);
$vars1 is now build this way
$vars1[row][column] contains the required value...
$vars2[column][row] contains the required value...
Maybe this helps:
?php
/**
mySQL - Database Class
**/
// Database Configuration Variables (local) //
define (DB_SERVER,localhost);
define (DB_USER,root);
define (DB_PASS,);
define (DB_DB,testfaq);
class Ms_db
{
/**
* Attributes of this class
**/
var $db_result = 0; // Stores the last result_id
var $db_id = 0; // Stores the current Database Connection ID
var $db_datasets = array(); // Stores the last read database-Set as hash
[colum-name][row_number]
/**
* Connect
* @author Marcel Schindler
* @param bool $persistant
* @return void
**/
function connect($persistant =
FALSE,$server=DB_SERVER,$user=DB_USER,$pass=DB_PASS,$database=DB_DB)
{
if (($persistant == FALSE))
{
$this-db_id = @mysql_connect($server,$user,$pass)
or $this-db_error(Failed to connect to
Database-Server,mysql_error());
}
else
{
$this-db_id = @mysql_pconnect($server,$user,$pass)
or $this-db_error(Persistant connection
failed,mysql_error());
}
@mysql_select_db($database) or $this-db_error(Failed to select
Database,mysql_error());
}
/**
* Query
* @param SQL-Query
* @param flip boolean
* @return array $dataset
**/
function query($sql,$flipped=on)
{
$db = $this-db_id;
$wert = array();
$count= 0;
if (!$db) $this-db_error(Keine Verbindung zur Datenbank);
$res = @mysql_query($sql) or $this-db_error(Fehlerhaftes
SQL-Statement,$sql.br.mysql_error());
while ($data = mysql_fetch_assoc($res))
{
$wert[$count] = $data;
$count ;
}
if ($flipped == on) $wert = $this-reorder_array($wert);
$this-db_datasets=$wert;
return $wert;
}
/**
* Close - closes the connection to the Database
**/
function close()
{
mysql_close();
}
/**
* Reorder_array
* @param: $array = array();
* @return: $array;
**/
function reorder_array($arr)
{
$wert = array();
// Array ist vom Typ $wert[Zeile][Spalte], soll aber als
$wert[Spalte][Zeile]
foreach ($arr as $sub)
{
while (list ($key1,$val1) = each ($sub))
{
$wert[$key1][] = $val1;
}
}
$this-db_datasets = $wert;
return $wert;
}
/**
* db_error - displays a well-formatted HTML-Page with the error-message
* @param string errormessage
* @param string mysql-error (optional)
**/
function db_error($message,$error='')
{
echo 'htmlheadtitle'.$message.'/title/headbody
bgcolor=#ee';
echo 'table align=center width=600 height=400
bgcolor=#ff';
echo 'trtdh1Error:/h1/td/trtrtd';
highlight_string($message);
if ($error!='')
{
echo 'brbrMySQL said:'.$error;
}
echo '/td/tr/table/body/html';
die();
}
/**
* NumRows
* Checks the Number of rows affected by the last query
* @return integer $number
**/
function numrows()
{
$number = @mysql_num_rows($this-db_result);
return $number;
}
/**
* DUMP - just dumps the current array (just for informal purposes
* @param VOID
**/
function dump()
{
$x = $this-numrows();
if ($x == 0 ) $this-db_error(There was no query placed
before,);
echo 'strongQuery DUMP/strongbrtable border=1
cellspacing=0 cellpadding=0 width=100%tr';
$wert = $this-db_datasets;
$spalten = array_keys($wert);
foreach ($spalten as $values)
{
$spaltenname[] = $values;
echo th bgcolor=\#00\font
color=\#ff\$values/font/th;
}
echo /tr;
for ($t = 0; $t = $x; $t )
{
if ($t % 2 == 0) echo tr bgcolor=\#dd\;
else echo tr bgcolor=\#ee\;
foreach($spaltenname as $values)
{
echo td vAlign=\top\.$wert[$values][$t]./td;
}
[PHP-DB] Re: Session data not being deleted on browser close.
Don't cross post such question... All you need to understand is how cookie is managed unless you are passing session id via URL. Read RFC2965 and RFC2964. You probably want to read netscape cookie spec also. -- Yasuo Ohgaki Youngie wrote: Why would my session data not be deleted after my browser is closed? I can set some session variables, close my browser, reopen them and the old values are still present, I can verify this by seeing that the file still containts my session data and values. Thanks John. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Retreiving data from a table
i'm guessing you use mysql, so you can use the mysql_insert_id to retrieve the most recent addition, then select on it $newid=mysql_insert_id(); $sql=select [the auto_increment field] from table1 where id='$newid'; Does that help? Morten Nielsen wrote: Hi, I got the following insert statement in my PHP code: $qid = db_query( INSERT INTO table1 ( Name ) VALUES ( '$frm[Name]' )); This is working like I want it to But in the table there is a column, which is AUTO_INCREMENT I would like to retreive this number right after I added the data Can anybody help me? Thanks, Morten -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Re: Insert data selected from drop down list into mySQL table
George, When you use: pre OPTION value=$prd$prd/OPTION /pre Why not use the Product ID in the VALUE tag, and use the Product NAME between the OPTION /OPTION tags? That way, the VALUE is what is selected, and you can then use an SQL INSERT or UPDATE command to push that into the destination table. One less check against the database, right? -Peter === EASY and FREE access to your email anywhere: http://Mailreader.com/ === -- 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] Re: Retrieving data woes
Andres Sulleiro [EMAIL PROTECTED] wrote in message
002001c18193$eacb4450$[EMAIL PROTECTED]">news:002001c18193$eacb4450$[EMAIL PROTECTED]...
I have a db table that has these columns:
id, date, time, title, body
some of the rows can have the same date and want to output the data
according to date. Like so:
dateA
timeA1, titleA1
bodyA1
timeA2, titleA2
bodyA2
dateB
timeB1, titleB1
BodyB1
In your SQL request, sort by date. When printing the output,
keep track of the previous date - only print the date if it
changes. Something like
$conn = mysql_pconnect($host, $usr, $pwd);
mysql_select_db(mydb);
$sql = SELECT id,date,time,title,body FROM mytable ORDER BY date DESC;
$res = mysql_query($sql);
$date = previous date;
while($row = mysql_fetch_array($res)) {
if ($date != $row[date]) {
$date = $row[date];
echo br$date;
}
echo p.$row[time]. .$row[title];
echo br.$row[body];
}
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[PHP-DB] Re: new data detect
Sommai Fongnamthip [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hi, I decide to make real time application (like stock market or online news). How could I know when new data arrive to table and retrieve them to display or update in web page? You can't, actually, short of using server-side push; what you can do is periodically requery the server; either reload the page every so often, or (better) use a flag file - set it when data is changed - and check it in the background with JavaScript, then reload the page only when it is set. -- 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]
