header ( ' Location .... with variable

[Ladbroke]

Hiya,

got a little prob with the syntax for
header ( ' Location .... with variable).
Usually examples are given with a complete URL, like:

..... header( 'Location: login/testsite/start.php' );

I'd like to use particular read out data from my database (URL)
(variable $result).


..... header( 'Location: $result' ); ?????

$result is not a string but an array
see below in code: $result=mysql_fetch_array($sqlrequest);

Thanx in anticipation!



Code

<?php

session_start ();

if(isset($_POST['password'])

AND
strcmp(trim($_POST['password']),'') != 0 )


$dbID=mysql_connect ('localhost', 'root') or die(mysql_error());

mysql_select_db('login', $dbID)or die(mysql_error());

$sqlZiel="SELECT destination FROM passwordrequest WHERE password = '101TR'";

//'101TR' (example)

$sqlrequest=mysql_query($sqldestination)or die(mysql_error());

$Ergebnis=mysql_fetch_array($sqlrequest);

mysql_close();

//print_r ($result);

if(isset($result))

{

header("Location: ($result)");

//header( 'Location: testsite/start.php' );

exit();
}

else

{

header( 'Location: http://localhost/login/IDfailure.php' );

exit();
}


?>

[Dormilich]

try

header('location: ' . $result[0]);
# or
header('location: ' . $result['destination']);

please use [ CODE ] tags when posting code

[Markus]

Single quotes will not parse variables unless they are concatenated, like the above post shows. Double quotes, however, will parse variables without having to concatenate them.