Re: [PHP] INSERT problem with MySQL/PHP

2001-06-19 Thread Ethan Schroeder

replace $result = mysql_db_query('item_db', $sql_query, $connection_id);
with $result = mysql_db_query('item_db', $sql, $connection_id); or just
mysql_query($sql);


- Original Message -
From: Todd A. Jacobs [EMAIL PROTECTED]
To: PHP General [EMAIL PROTECTED]
Sent: Monday, June 18, 2001 4:11 PM
Subject: [PHP] INSERT problem with MySQL/PHP


 PHP: 4.0.4pl1
 MySQL: 3.23.36-1

 I have the following code fragment, which uses the same connection
 parameters elsewhere to *successfully* to retrieve data from a database,
 so this doesn't appear to be a permissions problem with MySQL.

 However, nothing happens with inserts; no rows are added at all. All I get
 is a message form mysql_error() saying Query was empty.

 Can anyone tell me what might be wrong?

 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';

 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
 or die (No connection.\n);

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
 '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

 echo pre;
 echo \nDebugging output:\n;
 echo \tSQL: $sql\n;
 echo \tError: , mysql_error(), \n;
 echo \tLast insert ID: , mysql_insert_id(), \n;
 echo \tResult: $result\n;
 echo /pre\n;

 --
 Todd A. Jacobs
 CodeGnome Consulting, LTD





Re: [PHP] INSERT problem with MySQL/PHP

2001-06-19 Thread Todd A. Jacobs

On Mon, 18 Jun 2001, Todd A. Jacobs wrote:

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
   '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

Thanks to all who answered. It was a simple mistake, but hard for the eye
to catch: I needed the query to use $sql, not $sql_query.

The program now works properly. Thank you.

-- 
Todd A. Jacobs
CodeGnome Consulting, LTD




RE: [PHP] INSERT problem with MySQL/PHP

2001-06-19 Thread Jason Murray

 Can anyone tell me what might be wrong?
 
 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';
 
 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
   or die (No connection.\n);
 
 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
   '$question', '$question_type', NULL, NULL);
 
 $result = mysql_db_query('item_db', $sql_query, $connection_id);
  ^^

I daresay this will be the problem since the SQL is in $sql.

Jason



Re: [PHP] INSERT problem with MySQL/PHP

2001-06-18 Thread Tom Carter

Only a small little problem.. you called your sql query $sql and then in
your mysql_db_query line called it $sql_query..call them both the same name
and it should work!

HTH,Tom
- Original Message -
From: Todd A. Jacobs [EMAIL PROTECTED]
To: PHP General [EMAIL PROTECTED]
Sent: Tuesday, June 19, 2001 12:11 AM
Subject: [PHP] INSERT problem with MySQL/PHP


 PHP: 4.0.4pl1
 MySQL: 3.23.36-1

 I have the following code fragment, which uses the same connection
 parameters elsewhere to *successfully* to retrieve data from a database,
 so this doesn't appear to be a permissions problem with MySQL.

 However, nothing happens with inserts; no rows are added at all. All I get
 is a message form mysql_error() saying Query was empty.

 Can anyone tell me what might be wrong?

 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';

 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
 or die (No connection.\n);

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
 '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

 echo pre;
 echo \nDebugging output:\n;
 echo \tSQL: $sql\n;
 echo \tError: , mysql_error(), \n;
 echo \tLast insert ID: , mysql_insert_id(), \n;
 echo \tResult: $result\n;
 echo /pre\n;

 --
 Todd A. Jacobs
 CodeGnome Consulting, LTD





Re: [PHP] INSERT problem with MySQL/PHP

2001-06-18 Thread David Robley

On Tue, 19 Jun 2001 08:41, Todd A. Jacobs wrote:
 PHP: 4.0.4pl1
 MySQL: 3.23.36-1

 I have the following code fragment, which uses the same connection
 parameters elsewhere to *successfully* to retrieve data from a
 database, so this doesn't appear to be a permissions problem with
 MySQL.

 However, nothing happens with inserts; no rows are added at all. All I
 get is a message form mysql_error() saying Query was empty.

 Can anyone tell me what might be wrong?

 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';

 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
   or die (No connection.\n);

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
   '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

 echo pre;
 echo \nDebugging output:\n;
 echo \tSQL: $sql\n;
 echo \tError: , mysql_error(), \n;
 echo \tLast insert ID: , mysql_insert_id(), \n;
 echo \tResult: $result\n;
 echo /pre\n;

Your INSERT statement is contained in the variable $sql but you are using 
$sql_query in the mysl_db_query function?

-- 
David Robley  Techno-JoaT, Web Maintainer, Mail List Admin, etc
CENTRE FOR INJURY STUDIES  Flinders University, SOUTH AUSTRALIA  

   URA Redneck if your funeral has more pickup trucks than cars.



Re: [PHP] INSERT problem with MySQL/PHP

2001-06-18 Thread Developer

Try the following,

INSERT INTO questions(user_id,objective,question,question_type)
VALUES('$uawr_idx', '$objective', '$question', '$question_type')

Define the rows you want to insert into.

Hope this helps!

Regards,
Ray
- Original Message -
From: Todd A. Jacobs [EMAIL PROTECTED]
To: PHP General [EMAIL PROTECTED]
Sent: Monday, June 18, 2001 4:11 PM
Subject: [PHP] INSERT problem with MySQL/PHP


 PHP: 4.0.4pl1
 MySQL: 3.23.36-1

 I have the following code fragment, which uses the same connection
 parameters elsewhere to *successfully* to retrieve data from a database,
 so this doesn't appear to be a permissions problem with MySQL.

 However, nothing happens with inserts; no rows are added at all. All I get
 is a message form mysql_error() saying Query was empty.

 Can anyone tell me what might be wrong?

 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';

 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
 or die (No connection.\n);

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
 '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

 echo pre;
 echo \nDebugging output:\n;
 echo \tSQL: $sql\n;
 echo \tError: , mysql_error(), \n;
 echo \tLast insert ID: , mysql_insert_id(), \n;
 echo \tResult: $result\n;
 echo /pre\n;

 --
 Todd A. Jacobs
 CodeGnome Consulting, LTD




Re: [PHP] INSERT problem with MySQL/PHP

2001-06-18 Thread Miles Thompson

Todd,

Did you echo $sql and examine it?
Did you try the statement at the mysql console?
Does the list of values exactly match the field list?
Did you try the alternate syntax 
INSERT [LOW_PRIORITY | DELAYED] [IGNORE]
[INTO] tbl_name
SET col_name=expression, col_name=expression, ...
as in the MySQL docs?

I'm really curious about what $sql looks like when you echo it. I've been 
surprised more than once.

Miles Thompson

At 04:11 PM 6/18/01 -0700, Todd A. Jacobs wrote:
PHP: 4.0.4pl1
MySQL: 3.23.36-1

I have the following code fragment, which uses the same connection
parameters elsewhere to *successfully* to retrieve data from a database,
so this doesn't appear to be a permissions problem with MySQL.

However, nothing happens with inserts; no rows are added at all. All I get
is a message form mysql_error() saying Query was empty.

Can anyone tell me what might be wrong?

 $user_idx  = 1;
 $objective = 1;
 $question  = 'Foo';
 $question_type = 'singular';

 $connection_id = mysql_connect ('192.168.1.1', 'php', 'password')
 or die (No connection.\n);

 $sql = INSERT INTO question VALUES (NULL, $user_idx, $objective,
 '$question', '$question_type', NULL, NULL);

 $result = mysql_db_query('item_db', $sql_query, $connection_id);

 echo pre;
 echo \nDebugging output:\n;
 echo \tSQL: $sql\n;
 echo \tError: , mysql_error(), \n;
 echo \tLast insert ID: , mysql_insert_id(), \n;
 echo \tResult: $result\n;
 echo /pre\n;

--
Todd A. Jacobs
CodeGnome Consulting, LTD