Bug #13948 Updated: mysql_free_result giving error for valid $result
| From: | jpm@php.net | Date: | Tue, 06 Nov 2001 14:47:13 +0000 |
| Subject: | Bug #13948 Updated: mysql_free_result giving error for valid $result | ||
| References: | 1 | Groups: | php.dev |
| Request: | Send a blank email to php-dev+get-69697@lists.php.net to get a copy of this message | ||
ID: 13948
Updated by: jpm
Reported By: rohan.hawthorne@nt.gov.au
Old Status: Duplicate
Status: Bogus
Bug Type: MySQL related
Operating System: Red Hat Version 2.4
PHP Version: 4.0.4pl1
Previous Comments:
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[2001-11-05 22:36:48] rohan.hawthorne@nt.gov.au
Here is the code:
$SQL = "UPDATE tblAsset SET fldAssetId = 2, fldAssetCode = 'bc04495' WHERE fldAssetId
= 2";
$result = mysql_query("$SQL");
if (!$result) { echo("ERROR: " . mysql_error() . "$SQL"); }
mysql_free_result ($result);
Here is the error:
Warning: Supplied argument is not a valid MySQL result resource in
/var/www/html/itss/ProcessAsset.php on line 4
If an UPDATE doesn't produce a positive $result, why does the first error check (line 3) not
produce an error ?
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Edit this bug report at http://bugs.php.net/?id=13948&edit=1