mysql warning...

[phence]

i am getting this warning...

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/www/user/test/ya/creat_art.php</b> on line 35
Error 0 :

from this line of code...

	$result = mysql_query($query, $connection) or die("Error " . mysql_errno() . " : " . mysql_error());

if anyone could assist i'd really appreciate it!
thank-you!😃

[buddysal]

whats $query?

take out $connection...

$result = mysql_query("SELECT * FROM WHATEVER");

[phence]

$query = "SELECT * FROM categories WHERE Parent IS NULL";

need $connection to to connect to the database

[buddysal]

try this...

$user = "username";
$pass = "pass";
$db = "database";
$link = mysql_connect( "localhost", $user, $pass );
mysql_select_db( $db, $link );

$result = mysql_query("SELECT * FROM categories WHERE Parent IS NULL");

[phence]

yeah i pretty much have that.. heres the whole code

<SELECT name="categories">
<?php
// category functions
function categories ($id, $depth) {
	$query = "SELECT * FROM categories WHERE Parent IS NULL";
    $result = mysql_query($query);
    if (mysql_num_rows($result) < 1)
        return;
    while($row = mysql_fetch_array($result)) {
        $ID = $row['ID']; 
		$Name = $row['Name']; 
		categories2 ($ID, $Name, $depth, $query, $connection);        

    }
}
function categories2 ($ID, $Name, $depth, $query, $connection) {
	if (($result = mysql_query ($query, $connection)))
	{
    	while ($category = mysql_fetch_array($result))
		{

?>
        <option name="<?php echo $category['ID']; ?>"> 
<?php 
	echo($depth . $Name); 
?>
   </option>
<?php
    $depth = $depth . $Name . " > ";
    categories ($ID, $depth); 
	// contiue if statement
		}
	}         

	else
        echo "umm.. somthing messed up";
}

// db data
global $hostName;
global $username;
global $password;
global $databaseName;

$hostName = 'localhost';
$username = 'iuser';
$password = 'pass';
$databaseName = 'db';

// Open a connection to the DBMS
if (!($connection = @ mysql_pconnect($hostName, $username, $password)))
	showerror();

if (!mysql_select_db($databaseName, $connection))
	showerror();

// Run the Functions
categories("NULL", "");
?>
      </select>