Mysql Result Warnings

[Utherr12]

I made an account creation form with username mail and password. The form works and the php code too because i see the entries in my sql db... BUT it gives me warnings:Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 5 in D:\webserver\functions.php on line 9 Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 6 in D:\webserver\functions.php on line 10 .create_acc.php :

<?phpini_set('display_errors', 1);error_reporting(E_ALL);include("functions.php");$valid = create_acc(strtolower($_POST['Username']),$_POST['pass'],strtolower($_POST['Email']));if ($valid)	{echo "<p>Account succesfully created</p><br />";}	else	{echo "<p>Account name or Email already exists</p><br />";}?>

functions.php:

<?phpfunction create_acc($username,$password,$mail){	$phpDB = mysql_connect("ip","xxxx","xxx");	mysql_select_db("php",$phpDB);		$username_array = mysql_query("SELECT `name` FROM `test_table` WHERE `name` = '$username';",$phpDB);	$mail_array = mysql_query("SELECT `mail` FROM `test_table` WHERE `mail` = '$mail';",$phpDB);		$username_db = mysql_result($username_array,0);	$mail_db = mysql_result($mail_array,0);		if(($username_db!=$username)&&($mail_db!=$mail))	{		$date_joined = date("d/m/Y H:i:s");		$password_hash = hash('md5',$password,false);				$createAcc = mysql_query("INSERT INTO `test_table` (`name`,`mail`,`pw_hash`,`date_joined`) VALUES ('$username','$mail','$password_hash','$date_joined');",$phpDB);		return 1; // 1 == Success	  }		else return 0; // 0 == failed	mysql_close($phpDB);}?>

Even though account creation works, i'm getting those two Warnings. I didn't had any problems with this before i decided to integrate this into a function on a separate file.My sql db had only 1 entry before i created the account with 0's on every column for testing.I'm mentioning that those warnings only appear when account doesn't yet exist in DB ... if i try to make a duplicate account, no warning appears and the function returns 0.

[justsomeguy]

You're not checking if the query returned records, you're just assuming there's one there.

$result = mysql_query("SELECT `name` FROM `test_table` WHERE `name` = '$username';",$phpDB) or exit(mysql_error($phpDB);if ($row = mysql_fetch_assoc($result)){  // existing username found  $name = $row['name'];}else{  // existing username not found}