return query result from class

[vanpoeck]

Hi all!

I'm beginning to think I must be mentally disturbed 😉

Case:
Using a self made DB class, I want to get the query result sent back from this class and perform the classic while($row = mysql_fetch_object($myClass->getDataset()) on it.

But it always turns me back : "mysql_fetch_object : supplied argument is not a valid result resource". Indeed, the class seems to turn back #3 instead of the resource reference [#3].

All other operations I provide in the class work fine: adding, deleting, modifying records, and even getting a single record turned back (but in the latter case I do not need a dataset to be turned back from the class)

Can anyone help me out on this?

Code follows:

---------- Class snippet -----------
var $numresult;
var $dataset;

// user vars
var $username;
var $password;
var $email;

            // Set methods
function setDataset($result) {$this->dataset = $result;}

// Get methods
function getDataset() {return $this->dataset;}

// Get all users for table given in parameter $tablename
function getUsers($tablename = "uploaders")
{
	$this->setTablename($tablename);
	$SQL = "select Usr, Pwd, Email from $this->tablename";
	$result = mysql_query($SQL, $this->myConn) or QueryError($SQL);
	$this->setDataset($result);
	mysql_free_result($result);
}

---------- Calling code snippet -----------
$myClass = new theClass;
$myClass->getUsers();
while ($row=$myClass->getDataset()) { // Here is where it hurts :mad:
}
....

[bubblenut]

mysql_free_result($result) is saying to the mysql server that you no longer need to be dealing with that dataset so the link to it is closed. Skip this line and it should work.
HTH
Rob

[bad76]

Originally posted by vanpoeck
$row = mysql_fetch_object($myClass->getDataset()

Hi,
I think this is contrary to OO programming.
Maybe better if instead a dataset you get the row self.
You create a function called getNext($row), writing in $row the next row and returning false if db ends.

So your while is just

while(getNext($row))
{
}

This helps you?

[vanpoeck]

Doh!
Thanks Rob! Forgot the basics here:
$foo = "attribute value";
$myvar = $foo //pointer creation

And yes, bad76, I suppose you're right. I'm just a newbie to OO programming 😉

Thanks to you all!