PHP MYSQL echo not working - php

Question

Hey guys for some reason the code is not working its not echo the success which is weird it inserts the mysql info but just not echo the success if you guy's know the problem please post a fix or something thanks a lot guy's!
<?php
//Initiliaze Database connection
require("config.php");
//IMPORTANT STUFF
$username = mysql_real_escape_string(stripslashes($_POST["strUsername"]));
$password = mysql_real_escape_string(stripslashes($_POST["strPassword"]));
$pass1 = gen_token($password, $username);
$age = mysql_real_escape_string(stripslashes($_POST["intAge"]));
$dob = mysql_real_escape_string(stripslashes($_POST["strDOB"]));
$email = mysql_real_escape_string(stripslashes($_POST["strEmail"]));
$gender = mysql_real_escape_string(stripslashes($_POST["strGender"]));
$classid = mysql_real_escape_string(stripslashes($_POST["ClassID"]));
$eyecolor = mysql_real_escape_string(stripslashes($_POST["intColorEye"]));
$skincolor = mysql_real_escape_string(stripslashes($_POST["intColorSkin"]));
$haircolor = mysql_real_escape_string(stripslashes($_POST["intColorHair"]));
$hairid = $_POST['HairID'];
//Checks if Email has Already been used
$emailcheck = mysql_query("SELECT id FROM users WHERE Email='$email'") or die("status=Error&strReason=" . mysql_error());
if (mysql_num_rows($emailcheck) != 0) {
die("status=Taken&strReason=The email is already in used by another user.");
}
//Checks If Username has been Taken
$sql = mysql_query("SELECT * FROM users WHERE Username = '$username'") or die("status=Error&strReason=" . mysql_error());
if (mysql_num_rows($sql) !=0) {
die("status=Taken&strReason=The username is already in use by another character.");
} else {
//Sets Hairname & hairfile
switch ($hairid) {
//MALE HAIR
case 52:
$hairname = 'Default';
$hairfile = 'hair/M/Default.swf';
break;
case 55:
$hairname = 'Goku1';
$hairfile = 'hair/M/Goku1.swf';
break;
case 58:
$hairname = 'Goku2';
$hairfile = 'hair/M/Goku2.swf';
break;
case 64:
$hairname = 'Normal2';
$hairfile = 'hair/M/Normal2.swf';
break;
case 92:
$hairname = 'Ponytail8';
$hairfile = 'hair/M/Ponytail8.swf';
break;
//FEMALE HAIR
case 14:
$hairname = 'Pig1Bangs1';
$hairfile = 'hair/F/Pig1Bangs1.swf';
break;
case 18:
$hairname = 'Pig2Bangs2';
$hairfile = 'hair/F/Pig2Bangs2.swf';
break;
case 26:
$hairname = 'Pony2Bangs2';
$hairfile = 'hair/F/Pony2Bangs2.swf';
break;
case 83:
$hairname = 'Bangs2Long';
$hairfile = 'hair/F/Bangs2Long.swf';
break;
case 84:
$hairname = 'Bangs3Long';
$hairfile = 'hair/F/Bangs3Long.swf';
break;
}
$time = date("Y-m-d");
//Inserts Character Info into DB
$sql2 = mysql_query("INSERT INTO `users` (`Username`, `Password`, `Access`, `ActivationFlag`, `Age`, `Gender`, `Email`, `Level`, `Gold`, `Coins`, `Exp`, `ColorHair`, `ColorSkin`, `ColorEye`, `ColorBase`, `ColorTrim`, `ColorAccessory`, `DateCreated`, `UpgradeExpire`, `UpgradeDays`, `BankSlots`, `HouseSlots`, `BagSlots`, `HairID`, `HairFile`, `HairName`, `Permamute`, `Quests`, `Settings`, `Achievement`, `Country`, `AchievementID`, `CurrentServer`) VALUES ('$username', '$pass1', '0', '5', '15', '$gender', '$email', '1', '0', '0', '0', '$haircolor', '$skincolor', '$eyecolor', '0', '0', '0', '$time', '$time', '-1', '0', '20', '150', '$hairid', '$hairfile', '$hairname', '0', '00000000000000000000000000000000000000000000000000', '0', '0', 'US', '', 'Offline');") or die("status=Error&strReason=" . mysql_error());
//Selects New User ID
$sql3 = mysql_query("SELECT * FROM users WHERE Username='$username'") or die("status=Error&strReason=" . mysql_error());
$user = mysql_fetch_assoc($sql3) or die("status=Error&strReason=" . mysql_error());
$userId = $user['id'];
//Add's Starting Armor
switch ($classid) {
case 2:
$addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('2', '$userId', '1', 'ar', '1')");
break;
case 4:
$addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('4', '$userId', '1', 'ar', '1')");
break;
case 3:
$addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('3', '$userId', '1', 'ar', '1')");
break;
case 5:
$addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('5', '$userId', '1', 'ar', '1')");
break;
}
// ADDS DEFAULT WEAPON
$addweapon = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('1', '$userId', '1', 'Weapon', '1')" );
// ADDS USERS FRIEND LIST
$addfriends = mysql_query("INSERT INTO users_friends (userid, friends) VALUES ($userId, '')" );
//SUCCESS
echo "status=Success";
}
function gen_token($pass, $salt) {
$salt = strtolower($salt);
$str = hash("sha512", $pass.$salt);
$len = strlen($salt);
return strtoupper(substr($str, $len, 17));
}
?>

Answer 1

I count 4 INSERT commands that will ALWAYS occur (assuming successful). Is this correct, or are only some of them working?
Nothing jumps out as incorrect. You're storing the mysql_query result in variables, but not doing anything with them. Maybe try testing those values to see what's returned by MySQL?
You're also not testing to see if any of the POST'ed values are null, empty, or invalid. Always validate what's being POST'ed - for quick debugging, you could do a print_r($_POST);.