Sorry my mistake the array is $this->query_id the function is $query I still 
get the same results though.







>From: Rick Emery <[EMAIL PROTECTED]>
>To: 'Anzak Wolf' <[EMAIL PROTECTED]>, [EMAIL PROTECTED]
>Subject: RE: [PHP] MySQL arrays
>Date: Tue, 26 Mar 2002 09:23:11 -0600
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>how can you take a size of a function: $len = sizeof($this->query);
>and WTF (query is a function, not an array):  $tq = &$this->query[$q_id];
>
>-----Original Message-----
>From: Anzak Wolf [mailto:[EMAIL PROTECTED]]
>Sent: Tuesday, March 26, 2002 9:09 AM
>To: [EMAIL PROTECTED]
>Subject: [PHP] MySQL arrays
>
>
>I'm trying to do something a little different in my database class I have a
>method to do the query and store it in an array.
>
>function query ($s = "") {
>         $q = mysql_query($s,$this->database_connect_id);
>         if (!$q) {
>                 $tools->error(array("Query Resulted in NULL value"));
>                 return 0;
>         }
>         $len = sizeof($this->query);
>         $len++;
>         $this->query = array($len=>&$q);
>         mysql_free_result($q);
>         return $len;
>}
>
>function num_rows($q_id) {
>         // Set temp array to the selected query
>         $tq = &$this->query[$q_id];
>         $rows = mysql_num_rows($tq);
>         return $rows;
>}
>
>When I do a var_dump of the two key vars I get the following
>
>var_dump($this->query);
>array(1) {
>   [1]=>
>   resource(2) of type (mysql result)
>}
>var_dump($tq);
>resource(2) of type (Unknown)
>
>Here is the error that happens when I do the mysql_num_rows
>
><br>
><b>Warning</b>:  2 is not a valid MySQL result resource in
><b>/var/www/includes/database_class.php</b> on line <b>77</b><br>
>
>any ideas as to what I can do to fix this so that I can store query for
>later use?
>
>-Jim
>
>
>
>
>
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