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

ID: 10389 Updated by: cnewbill Reported By: pmarks@employees.org Old-Status: Open Status: Feedback Bug Type: MySQL related PHP Version: 4.0.4pl1 Assigned To: Comments: 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 Previous Comments: --------------------------------------------------------------------------- [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. --------------------------------------------------------------------------- ATTENTION! Do NOT reply to this email! To reply, use the web interface found at http://bugs.php.net/?id=10389&edit=2