ID: 10389
Updated by: cnewbill
Reported By: [EMAIL PROTECTED]
Old-Status: Bogus
Status: Feedback
Bug Type: MySQL related
PHP Version: 4.0.4pl1
Assigned To:
Comments:
My bad...it would not return false. It's been a long long day.
But it would return false if the table jobsc did not exist or the database ue did not
exist.
-Chris
Previous Comments:
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[2001-04-18 20:26:15] [EMAIL PROTECTED]
Your query is not returning any rows, and as such mysql_query returns false. False of
course is not a valid mysql result and thus the "Warning".
If you are 100% certain the query returns results then that would be a problem,
otherwise this is not a bug.
Please check this query from the mysql command line and reopen if there really is a
problem.
-Chris
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[2001-04-18 19:45:58] [EMAIL PROTECTED]
Here is the code I am using that produces the error:
<?php
$db = mysql_connect("localhost");
mysql_select_db("ue", $db);
$result = mysql_query("SELECT * FROM jobsc");
$rows = mysql_num_rows($result);
?>
on the last line is where the error is generated. In
any browser, I see:
Warning: Supplied argument is not a valid MySQL result
resource ...
My setup contains apache 1.3.19 with the static PHP 4.0.4pl1
module compiled in. My configure lines for PHP are:
./configure --with-apache=/home/apache_1.3.19
--with-mysql=/usr/local/mysql
/usr/local/mysql is the install directory for mysql
That is all the information I believe I have, and I'm
relatively sure that PHP is not crashing, per say.
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