Getting a blank page with PHP&MySQL - php

Question

I'm getting a blank page with this code:
session_start();
include "config.php";
$af = $_GET['id'];
database_connect();
$query2 = "SELECT * FROM friends WHERE usr1 = '".$id."' AND usr2 = '".$af."'";
$result2 = mysql_query($query2) or die(mysql_error());
while ($row2 = mysql_fetch_assoc($result2)) {
if($row2['id']){
echo "<script type='javascript'>alert('You are already friends with this person.');</script>";
header('Location: profile.php?id="'.$af.'"');
}else{
mysql_query("INSERT INTO friends (usr1, usr2)
VALUES ('".$id."', '".$af."')") or die(mysql_error());
echo "<script type='javascript'>alert('You two are friends now!');</script>";
header('Location: profile.php?id="'.$af.'"');
};
};
This is the config.php (i changed the variables here though)
$h = "localhost";
$u = "user";
$p = "pass";
$d = "datab";
$sql = 'SELECT id FROM craffyposts limit '.($page*$eachPage).','.$eachPage;
$sql_count = 'SELECT id FROM craffyposts';
function database_connect(){
global $h, $d, $u, $p;
$link = #mysql_connect("$h","$u","$p");
$sql_error = mysql_error();
if (!$link) {
echo "Connection with the database couldn't be made.<br>";
echo "$sql_error";
exit;
}
if (!#mysql_select_db("$d")) {;
echo "The database couldn't be selected.";
exit;
}
return $link;
}
if($_SESSION['usrid']){
database_connect();
$query = mysql_query("SELECT * FROM craffyusers WHERE id='" .$_SESSION['usrid']. "' ") or die (mysql_error());
while ($obj = mysql_fetch_object($query)) {
$id = htmlspecialchars($obj->id);
$username = htmlspecialchars($obj->username);
$email = htmlspecialchars($obj->email);
$realname = htmlspecialchars($obj->name);
$srvrid = htmlspecialchars($obj->serverid);
$propic = htmlspecialchars($obj->profilepic);
};
};
What's the issue here?

Answer 1

because there will 0 or 1 result, you can remove the while clause:
$row2 = mysql_fetch_assoc($result2);
if($row2 && $row2['id']){
echo "<script type='javascript'>alert('You are already friends with this person.');</script>";
header('Location: profile.php?id="'.$af.'"');
}else{
mysql_query("INSERT INTO friends (usr1, usr2)
VALUES ('".$id."', '".$af."')") or die(mysql_error());
echo "<script type='javascript'>alert('You two are friends now!');</script>";
header('Location: profile.php?id="'.$af.'"');
};

Answer 2

Add this lines at the beginning of the script
error_reporting(E_ALL);
ini_set("display_errors", 1);
The error will appear.