Re: [PHP4BETA] lates libzend update did bad things

At 14:20 03/09/99 , Thies C. Arntzen wrote:

On Fri, 3 Sep 1999, Zeev Suraski wrote: At 10:45 03/09/99 , Thies C. Arntzen wrote:

zeev, the lates update broke something... <? $conn=mysql_connect("127.0.0.1"); for ($inx = 0; $inx < 10; $inx++) {

    $r = mysql_db_query("test", "select * from phptest");
    $array = mysql_fetch_array(&$r);

// mysql_free_result($r); } echo "-----------------------end--------------\n";flush(); ?> used to work - now says: PHP Warning: 2 is not a MySQL result index in mysql.php3 on line 6

Well, why are you sending the result handle by reference? My API patch doesn't make miracles, it keeps the existing C code in a way it would work nicely, but not with weirdo stuff like that :) To avoid these errors, you would have to actually modify the C code to be PHP 4.0 style (which I will today, but I'll only do the MySQL module). sorry, that was a leftover.. it seems to work without the '&' - but have a look at #2230 - i have the same problem with some oracle-php-code i've written. Basically, mysql_fetch_array() expects to receive a copy of the result handle, and it destroys it when it's done with it (transparently, mysql_fetch_array() was never designed to do it). Since you now send it the actual handle itself and not a copy - the handle that's destroyed is the only handle, and the object is dead by the time PHP wants to use it. hmm got you - i will have a deeper look why my code stopped working this morning - again, i'm not the only one (#2230) - so some thing has changed;-)

Basically, if you take any resource by reference, and you convert_to_long() it - the resource goes bye-bye. With PHP 4.0, you shouldn't call convert_to_long() on an expected resource, but instead, check whether the type is IS_RESOURCE, bail out if it isn't, and just use it as-is. Zeev -- Zeev Suraski <zeev@zend.com> http://www.zend.com/ For a PGP public key, finger bourbon@netvision.net.il