Small problem, Not outputting. *SOLVED*

[xyn]

Hio,
I'm having a small problem with PHP, he seems to be getting
on my nerves tonight, well the thing is I have a avatar for
my members, yet I wanted to check if they ahve an avatar
or give them the default avatar.

So I check mysql as you do, and PHP doesn't seem to output
anything :/ could someone check my code, thx.

[code=php:0]<?PHP include("session.php");
$SQLB = mysql_query("SELECT filename FROM avatar WHERE user='{$_SESSION['username']['usr_user']}'");
while( $Data = mysql_fetch_array( $SQLB, MYSQL_NUM )){
$NUMB = mysql_num_rows( $SQLB );
if( !$NUMB ){
$avatar="avatar.PNG";
}else{
$avatar = $Data[0];
}
}
?>
<p>text here<br>
<img border="0" src="./global/images/private/<?PHP echo $avatar; ?>" alt="Avatar"></p>[/code]

[Barand]

If there is no data found, code inside the while loop is not executed

[xyn]

How could i fix it?

[Barand]

[code]<?php
include("session.php");
$SQLB = mysql_query("SELECT filename FROM avatar WHERE user='{$_SESSION['username']['usr_user']}'");
$NUMB = mysql_num_rows( $SQLB );
if( !$NUMB ){
    $avatar="avatar.PNG";
}else{
    $Data = mysql_fetch_array( $SQLB, MYSQL_NUM );
    $avatar = $Data[0];
}
?>
<p>text here<br>
<img border="0" src="./global/images/private/<?php echo $avatar; ?>" alt="Avatar"></p>
[/code]

[xyn]

worked. thanks :]. just confused me! thanks again.