Resource id #2 Error

Any Ideas

[brant]

The following queary is giving me problems

function checkID($ID) {
$link = open_npc();
$query = "SELECT status
FROM guards
WHERE ID='$ID'";
$result = mysql_query($query);
mysql_close($link);
if ($result['status'] > 0) return ture;
else return false;
} // end function checkID()

When I just return the results it gives me the following: Resource id #2 Error

[mincklerstraat]

you can get more info about this if you put this after your line

$result = mysql_query($query);

:

if(!$result) echo mysql_error(); 

when getting your script to upload to your live server, erase this line or comment it out.

[brant]

I placed it but it didn't do anything.

[brant]

It seems that the sql queary is not returning any inforamtion. I believe I am connected to the MySQL server but I am not sure.

[coopster]

Welcome to WebmasterWorld, brant.

Some things to check ...
Are you sure you are connecting to the database? Check your $link variable.
Dump the query to the browser to check the validity.
You have "true" spelled incorrectly.

function checkID($ID) { 
 $link = open_npc(); 
 $query = "SELECT status FROM guards WHERE ID='$ID'"; 
 exit($query); // dump the query to see what's wrong ... 
 $result = mysql_query($query); 
 mysql_close($link); 
 if ($result['status'] > 0) { 
   return true; 
 } else { 
   return false; 
 } 
} // end function checkID()