What is the problem? Please fix.

Liquify

$reporterarray = mysql_fetch_array(mysql_query("SELECT FROM Reports WHERE Username='$reporter'"));
$reportee = $qr['UserReported'];
$reporteearray = mysql_fetch_array(mysql_query("SELECT FROM Reports WHERE Username='$reportee'"));

Error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/socialli/public_html/Administration/ViewReports.php on line 28

NogDog

It means that MySQL was unable to process your query for some reason, so mysql_query() returned Boolean false instead of a query result resource. Instead of jamming everything into one line, try some defensive coding, instead.

$sql = "SELECT * FROM `Reports` WHERE `Username`='".mysql_real_escape_string($reporter)."'";
$result = mysql_query($sql);
if($result == false) {
  throw new Exception("Query failed:\n".mysql_error()."\n$sql"); // or however else you like to handle errors
}
$reporterarray = mysql_fetch_array($result);

dalecosp

And please, PLEASE stop using the MySQL extension ... please? 🙂

sneakyimp

Liquify;11034535 wrote:
$reporterarray = mysql_fetch_array(mysql_query("SELECT FROM Reports WHERE Username='$reporter'"));
$reportee = $qr['UserReported'];
$reporteearray = mysql_fetch_array(mysql_query("SELECT FROM Reports WHERE Username='$reportee'"));

Error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/socialli/public_html/Administration/ViewReports.php on line 28

Yes this is very bad code:
use the code formatting tags on this forum
stop using the mysql extension (see my signature)
check whether your query returns an error or not
check whether your query returns a result before you call mysql_fetch_array on it.