Script error

[SilverDragonTears]

I don't know how to get this to work with the new script. I've been trying since last night.

        $query = "SELECT * FROM ".$GLOBALS['prefix']."tabs WHERE username = '".$GLOBALS['username']."'";
	$result = runquery($query);
	$num = mysql_numrows($result);

[SilverDragonTears]

Ok here is what I have so far and it isn't working LOL

function gettabs() {
$formcontent = "";
$data = $adopts->select("tabs", array("username", "name"), "username = '{$username}'")->fetchObject();
$num = count($data);

	$i=0;
	while ($i < $num) {
		$tabname=@mysql_result($data, $i,"name");
		$formcontent = $formcontent."<option value='myadopts.php?tab=".$tabname."'>".$tabname."</option>";
		$i++;
	}
	return $formcontent;
}

[fadillzzz]

Don't mix the old mysql_* functions with PDO. The two are very different.

PHP Code:

$result = $adopts->select(...);
while ($row = $result->fetchObject())
{
    echo $row->name;
} 


[SilverDragonTears]

ok now I have this.. and still not working. Bear with me.

function gettabs() {
$formcontent = "";
$result = $adopts->select("tabs", array("username", "name"), "username = '{$username}'")->fetchObject();

	$i=0;
	while ($row = $result->fetchObject()) {
		$tabname=@mysql_result($data, $i,"name");
		$formcontent = $formcontent."<option value='myadopts.php?tab=".$tabname."'>".$tabname."</option>";
		$i++;
	}
	return $formcontent;
}

[Hall of Famer]

First of all, you are having an issue Id refer to as 'variable scope'. You are defining a function gettab() without passing any arguments, while the variable $username is undefined as a local variable inside your function. You either have to declare $username as global variable, or pass it as an argument to your function. Read this manual from PHP for reference:
http://php.net/manual/en/language.variables.scope.php

On the other hand, you still use mysql_result() which no longer works with this new script as we are incorporating PDO. I thought Fadillzzz already explained clearly to you not to use mysql functions anymore... Now change this line:

PHP Code:

$tabname=@mysql_result($data, $i,"name"); 


to

PHP Code:

$tabname=$row->name; 


And now the $tabname will not be malfunctioning. However I strongly recommend you to remove all lines associated with $i since its totally unnecessary.

[SilverDragonTears]

Ok another thing. I'm suppose to add this...

 AND ".constant("PREFIX")."owned_adoptables.tab = '{$tab}'

to the myadopts.php query... but when I do it still doesn't list the tabs.

Is this correct for functions.php?

function gettabs() {
$formcontent = "";
$result = $GLOBALS['adopts']->select("tabs", array("username", "name"), "username = '{$username}'");

	while ($row = $result->fetchObject()) {
		$tabname=$row->name; 
		$formcontent = $formcontent."<option value='myadopts.php?tab=".$tabname."'>".$tabname."</option>";
	}
	return $formcontent;
}

[Hall of Famer]

Umm are you trying to select from multiple tables? If so, use the method join() before calling select().

[SilverDragonTears]

I figured it out :D Thank you for the help!

[SilverDragonTears]

Hmm... this pulls up everyone's tabs instead of just the logged in user

$result = $GLOBALS['adopts']->select("tabs", array("username", "name"), "username = '{$username}'");

i see why but I don't know how to fix it.


heh.

$result = $GLOBALS['adopts']->select("tabs", array("username", "name"), "username = '".$GLOBALS['username']."'");

Seems to do the trick ;)

[Hall of Famer]

Of course the old script wont work, didnt I tell you before that you need to worry about the variable scope issue? What is $username in your function? If you neither pass it as an argument to your function tab(), nor declare it as global variable, it will be considered an undefined variable.