CHANGE:
> $len = sizeof($this->query_id);
> $len++;
> $this->query_id = array($len=>&$q);
> mysql_free_result($q);
> return $len;
TO:
$this->query_id[] = $q;
return sizeof($this->query_id);
second:
$q is a resource, that is, a pointer variable. When you execute
mysql_free_result($q), you free-up the memory that it pointed to.
Therefore, when you attempt to recall its value, it is pointing to memory
that you no longer own. Hence, it is not a valid resource. Therefore, do
not free the resource. When the script completes, the memory will
automatically be de-allocated. Also, when you use the phrase "&$q", you are
creating a pointer to a pointer; is that what you want?
>-----Original Message-----
>From: Anzak Wolf [mailto:[EMAIL PROTECTED]]
>Sent: Tuesday, March 26, 2002 9:09 AM
>To: [EMAIL PROTECTED]
>Subject: [PHP] MySQL arrays
>
>
>I'm trying to do something a little different in my database class I have a
>method to do the query and store it in an array.
>
>function query ($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_id);
> $len++;
> $this->query_id = array($len=>&$q);
> mysql_free_result($q);
> return $len;
>}
>
>function num_rows($q_id) {
> // Set temp array to the selected query
> $tq = &$this->query_id[$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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