php code problem

[einSTein]

i have a problem in this code , this code must check the users in the db first then if the user does not exist it will make a record in the db and if the user exist it will return back Username is already in use , my problem is that every time i enter a user name even it is in the db it makes a new user with the same username

$sql = mysql_query("SELECT * FROM login");
$flag=0;
while ($r = mysql_fetch_array($sql)) 
{ 
if ($r['user'] == $_POST['user']) 
{
$flag=1;
//searching for the user ib the db
echo '<font color="#FF0000">';
echo "* Username is already in use";
echo '</font>';
echo "<br>";
}
else 
{
//entering the job seeker data in the user table
$sql="INSERT INTO user (name, Addr, Nation, dob, Gen, cnt, mar, mil, phone, mobile, mail, user)
VALUES
('$fn','$_POST[add]','$_POST[nat]','$dob','$_POST[gg]','$_POST[cnt]','$_POST[Mar]','$_POST[Mil]','$_POST[ph]','$_POST[mo]','$_POST[mail]','$_POST[user]')";
//entering the user and pass in the login table
$sql2="INSERT INTO login (user, pass, type)
VALUES
('$_POST[user]','$_POST[pass]','$_POST[type]')";
if (!mysql_query($sql,$con))
    {
    die('Error: ' . mysql_error());
    }
    echo "<br>";
if (!mysql_query($sql2,$con))
    {
    die('Error: ' . mysql_error());
    }
    echo "<br>";
echo "<br>Thanks for registration $fn ";
exit();
} 

[byron3@earthlink]

$flag = 0;

$query = "SELECT * FROM login where user ='" . $_POST['user']. "'";
$result = $sql = mysql_query( $query );

if ( $result ) {
    $flag=1;
    echo '<font color="#FF0000">';
    echo "* Username is already in use";
    echo '</font>';
    echo "<br>";
} else {

// Do your inserts here

} 

[einSTein]

that results that it prints already in use even when the db is empty

[byron3@earthlink]

Do a var_dump on $result and tell me what it outputs.

[einSTein]

write me the syntex

[einSTein]

plz

[byron3@earthlink]

$query = "SELECT * FROM login where user ='" . $_POST['user']. "'";
echo $query; // Lets take a look @ your actual query

$result = $sql = mysql_query( $query );
var_dump( $result ); // Lets take a look@ your returned resultset 

[einSTein]

this is the result of echo $query
SELECT * FROM login where user ='mariam'
and this is the result of the var_dump
resource(2) of type (mysql result)

[byron3@earthlink]

$result = $sql = mysql_query( $query );
var_dump( $result ); // Lets take a look@ your returned resultset 

while ( $r = mysql_fetch_assoc( $result ) { // lets see whats in these two results
    var_dump( $r );
} 

[einSTein]

error in that line
while ( $r = mysql_fetch_assoc( $result )
error message
Parse error: parse error in

[byron3@earthlink]

while ( $r = mysql_fetch_assoc( $result )) { // lets see whats in these two results
    var_dump( $r );
} 

Sorry I left out a closing )

[einSTein]

it results SELECT * FROM login where user ='mariam'
& resource(2) of type (mysql result)

[byron3@earthlink]

$result = mysql_query( $query );
var_dump( $result ); // Lets take a look@ your returned resultset 

while ( $r = mysql_fetch_assoc( $result ) { // lets see whats in these two results
    var_dump( $r );
} 

Einstein try this one more time. the problem was in "$result = $sql = "