RE: [PHP] MySQL arrays
| From: | Rick Emery | Date: | Tue, 26 Mar 2002 15:23:11 +0000 |
| Subject: | RE: [PHP] MySQL arrays | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-90196@lists.php.net to get a copy of this message | ||
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[$q_id];
-----Original Message-----
From: Anzak Wolf [mailto:anzak@hotmail.com]
Sent: Tuesday, March 26, 2002 9:09 AM
To: php-general@lists.php.net
Subject: [PHP] MySQL arrays
I'm trying to do something a little different in my database class I have a
method to do the query and store it in an array.
function query ($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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