[RESOLVED] implode error

[Desbrina]

Warning: implode() [function.implode]: Bad arguments. in /usr/home/desbrina/public_html/mt/users/master.php on line 72
UPDATE user_inventory set status = 'mastered' WHERE ID IN()
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1

I get that error with this code

$sql = "UPDATE user_inventory set status = 'mastered' WHERE ID IN(".implode(', ', $ids).")";
$result = mysql_query($sql) or echo($sql."<br />".mysql_error());
print_r($ids);

IDs are entered via

$ids = array();
			while($row = mysql_fetch_array($sql)) {
				$ids[] = $row['ID'];
			}

I'm a bit confused since it was working fine yesterday

[laserlight]

The reason is that the $ids array is empty, so you end up with "IN ()", which is invalid SQL. You need to check for this array condition.

[Desbrina]

if i print it after its set, so after

$ids = array();
         while($row = mysql_fetch_array($sql)) {
            $ids[] = $row['ID'];
         }
print_r($ids);

Then it shows the IDs fine

But if i print it after the first piece of code then its blank. I changed the variable to session and it works, so the variable seems to be getting lost when the page is reloaded

[Weedpacket]

So you're saying

$ids = array();
         while($row = mysql_fetch_array($sql)) {
            $ids[] = $row['ID'];
         }
print_r($ids); 

works but

$ids = array();
            while($row = mysql_fetch_array($sql)) {
                $ids[] = $row['ID'];
            } 
$sql = "UPDATE user_inventory set status = 'mastered' WHERE ID IN(".implode(', ', $ids).")";
$result = mysql_query($sql) or echo($sql."<br />".mysql_error());
print_r($ids); 

Doesn't and even the print_r() is empty?