RE: [PHP-DB] UPDATE table problem
| From: | Rick Emery | Date: | Mon, 15 Oct 2001 17:04:20 +0000 |
| Subject: | RE: [PHP-DB] UPDATE table problem | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-13337@lists.php.net to get a copy of this message | ||
The reason you're getting "Resource = id#2" is because you've used $result
from a previous statement like:
$result = mysql_query()
$result is simply an identifier. To determine WHAT it identifies, you must
extract the rows to which it
refers, i.e.:
$result = mysql_query($some_query) or die(mysql_error());
while( $row = mysql_fetch_array($result) )
{
$update_query = "update guestbook set last_access=3Dnow() where =
guest_id='$result'";
mysql_query($update_query) or die(mysql_error());
// do other stuff
}
-----Original Message-----
From: George Lioumis [mailto:glioumis@adacom.com]
Sent: Monday, October 15, 2001 8:31 AM
To: php-db@lists.php.net
Subject: [PHP-DB] UPDATE table problem
Hi, everyone, again,
Consider the following piece of code:
-------------------------
$update_query =3D "update guestbook set last_access=3Dnow() where =
guest_id=3D'$result'";
$update_query =3D stripslashes($update_query);
echo $update_query;
echo "<BR><BR>";
$updated =3D mysql_query($update_query) or die(mysql_error());
if ($updated)
{
.............etc etc
}
--------------------------
When ran, it returns:
update guestbook set last_access=3Dnow() where guest_id=3D'Resource =
id#2'
The problem is that if($updated) returns TRUE, but the last_access field =
is not updated at all !!! :(((
Any ideas as far as the $update_query is concerned???
George