PHP 4.0 Bug #8831 Updated: a script works on 2 servers I use, and doesn't work on a third one

ID: 8831 User Update by: mage@diablo.hu Status: Open Bug Type: MySQL related Description: a script works on 2 servers I use, and doesn't work on a third one Okay, I have found a note about mysql_connect() in the manual: "In case a second call is made to mysql_connect() with the same arguments, no new link will be established, but instead, the link identifier of the already opened link will be returned." That can explain what is wrong in my script. However, in older php versions, I could do connect again (and get same identifier, I checked), and close it as many times as I connected. Now I have to upgrade all my older projects? Previous Comments: --------------------------------------------------------------------------- [2001-01-21 20:46:18] mage@diablo.hu Dear Support, One of my project were must go to another server, and I got a trouble with my functions. I try to write a short code to explain it. This script works well on my computer, windows 98 se, php 4.0.3. Also works on debian a server, php 4.0.3. There is another debian server (that was php 4.0.3 too), and this script crashes on it. I tried to reinstall mysql, php (the new 4.0.4pl), apache, but that did not help. I show a short code: ------ <?php $base = mysql_connect('localhost','kepes','***'); // I have a password of course... mysql_select_db('KEPES'); $rows = mysql_query("SELECT * FROM ARTICLE_STATE ORDER BY article_state_id",$base); test(); $rows = mysql_query("SELECT * FROM ARTICLE_STATE ORDER BY article_state_id",$base); mysql_close($base); function query_more($query,$base) { $result = array(); $data = mysql_query($query,$base); while ($row = mysql_fetch_array($data)) { $result[] = $row; } // if ($result) { // mysql_free_result($data); // } return $result; } function test() { $base = mysql_connect('localhost','kepes','***'); mysql_select_db('KEPES'); $rows = query_more("SELECT * FROM ARTICLE_STATE ORDER BY article_state_id",$base); mysql_close($base); } ?> ----- That outputs nothing on my computer, nothing on the first debian server. That's ok. But this outputs on the server where I'd use it the below: "Warning: Supplied argument is not a valid MySQL result resource in /home/kepes/public_html/test/test5.php4 on line 16 Warning: 1 is not a valid MySQL-Link resource in /home/kepes/public_html/test/test5.php4 on line 10" I figured out this is if I use at least 2 mysql connections same time. Another interest thing, if I comment out the: --- while ($row = mysql_fetch_array($data)) { $result[] = $row; } --- part of script, this also works on the 2nd server (but I can't use that way of course). Because my db-scripts are based upon methods like above, I would like to know what the problem is. I include some php info. PHP Version 4.0.5-dev System Linux ditto 2.2.17 #1 Mon Dec 4 23:29:54 CET 2000 i686 unknown Build Date Dec 10 2000 Configure Command './configure' '--with-mysql' '--with-apxs' Server API Apache Virtual Directory Support disabled Configuration File (php.ini) Path /usr/local/lib ZEND_DEBUG disabled Thread Safety disabled --------- mysql MySQL Support enabled Active Persistent Links 0 Active Links 0 Client API version 3.23.22-beta MYSQL_INCLUDE MYSQL_LFLAGS MYSQL_LIBS Directive Local Value Master Value mysql.allow_persistent On On mysql.max_links Unlimited Unlimited mysql.max_persistent Unlimited Unlimited ----- Thank You, Mage --------------------------------------------------------------------------- Full Bug description available at: http://bugs.php.net/?id=8831