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: --------------------------------------------------------------------------- [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 --------------------------------------------------------------------------- [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. --------------------------------------------------------------------------- ATTENTION! Do NOT reply to this email! To reply, use the web interface found at http://bugs.php.net/?id=10389&edit=2 -- PHP Development 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]