Returning a variable from a function problem

I have a form where the user inputs information - the code below is the error checking for one of the fields. The first IF statement just checks that the filed is not empty and works fine. Then I check a function I made to see if the user is already in a mysql database and the function returns 0, 1, or 2 depending on the various conditions. The first elseif works as I would expect - but it seems that the next two elseif's are just being ignored. Could someone possibly explain what I have wrong here. Thanks if (empty($form["name"])) { $error["name"] = "*"; $message["name"] = "Required field!"; } elseif($duplicate = duplicate_name($form["name"]) == 0) { $error["name"] = "*"; $message["name"] = "Screen name already used!"; . } elseif ($duplicate == 1) { $error["name"] = "*"; $message["name"] = "Database error - Please try again in a few moments!"; } elseif ($duplicate == 2) { $error["name"] = "*"; $message["name"] = "Server error - Please try again in a few moments!"; } function duplicate_name($name) { if (!($connect = @mysql_connect($serverhost,$serveruser,$serverpass))) { **** These are all defined above but not included here. $server_error = "2"; } if (!@mysql_select_db($databasename)) { $server_error = "1"; } if($server_error) { $duplicate = $server_error; } else { $query = "SELECT name FROM $tablename WHERE name = '$name'"; if(!$query) { $duplicate = 1} else { $result = mysql_query($query); $line = mysql_fetch_array($result); if($line['name'] == $name) { $duplicate = 0; } mysql_free_result($result); mysql_close($connect); } } echo $duplicate; ***** This shows that the variable $duplicate is assigned 0, 1, or 2 depending on the what conditions were met. return($duplicate); }