Warning: mysql_fetch_array() ?

why do this code bellow give me this error ?

<?php
require_once("global.php");

$sql = "SELECT id FROM users ORDER BY today DESC LIMIT 1";
$result = mysql_fetch_array($sql);
$update_user = mysql_query("update users set featured = featured + 1 where userid = " . $result["id"] . " 

limit 1");
      

?> 

PHP:

 

You are missing a mysql_query:

$sql = "SELECT id FROM users ORDER BY today DESC LIMIT 1";
mysql_query($sql);
$result = mysql_fetch_array($sql);

PHP:

 

ok, so now it looks like this and still errors me :/

<?php
require_once("global.php");

$sql = "SELECT id FROM users ORDER BY today DESC LIMIT 1";
mysql_query($sql);
$result = mysql_fetch_array($sql);
$update_user = mysql_query("update users set featured = featured + 1 where userid = " . $result["id"] . "

limit 1");
     

?>

PHP:

 

<?php
require_once("global.php");

$sql = mysql_query("SELECT id FROM users ORDER BY today DESC LIMIT 1");
$result = mysql_fetch_array($sql);
$update_user = mysql_query("update users set featured = featured + 1 where userid = " . $result["id"] . "

limit 1");

?>

PHP:

 

still get the error :/

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /test.php on line 5

 

there must be something wrong with your query then.

<?php
require_once("global.php");

$sql = mysql_query("SELECT id FROM users ORDER BY today DESC LIMIT 1") or die (mysql_error());
$result = mysql_fetch_array($sql);
$update_user = mysql_query("update users set featured = featured + 1 where userid = " . $result["id"] . " limit 1");

?>

PHP:

That'll tell you the mysql error.

 

I knew something was missing in my fix... sorry

 

aw.... i had spelled on of the collumns wrong >.< i should learn to read...

anyhow, thank you all for the help!