Bug #10389 Updated: mysql_query(); returns an invalid result resource

ID: 10389 Updated by: sniper Reported By: pmarks@employees.org Old-Status: Feedback Status: Closed Bug Type: MySQL related Operating system: PHP Version: 4.0.4pl1 Assigned To: Comments: Can't reproduce. If still problem with latest CVS, reopen. --Jani Previous Comments: --------------------------------------------------------------------------- [2001-04-19 17:40:19] cnewbill@php.net This appears to be related to 4.0.4pl1. I can reproduce it on a 4.0.4pl1 box, but not anything newer. Can you try 4.0.5 or a CVS snapshot? http://snaps.php.net/ -Chris --------------------------------------------------------------------------- [2001-04-19 16:42:54] pmarks@employees.org I checked what you said with mysqladmin and mysql. The database "ue" does exist, as does the table "jobsc." Thanks for saying something about the table being empty, as it was. I fixed this problem (I'd had it before, but i just did this new script by itself to make sure) and added 1 entry to the table "jobsc." PHP4 still returns the same warning as before on the "SELECT * FROM jobsc" and in the mysql console, executing that instruction returns the one and only row in the table jobsc. --------------------------------------------------------------------------- [2001-04-18 20:29:19] cnewbill@php.net 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 --------------------------------------------------------------------------- [2001-04-18 20:26:15] cnewbill@php.net 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] pmarks@employees.org 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. --------------------------------------------------------------------------- The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online. ATTENTION! Do NOT reply to this email! To reply, use the web interface found at http://bugs.php.net/?id=10389&edit=2