mysql_fetch_array error not understood

[netfrugal]

Below is how I generate a count query of records in my customers table:

<? $count = mysql_query("SELECT COUNT(*) FROM customers WHERE status = '1'");
 while ($row = mysql_fetch_array($count)) {
	   $count = $row['COUNT(*)'];
	   }
	   ?>
<? echo $count; ?> Active Customers

The count does return the proper number of records, but I get this error in the logs:

PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /customers.php on line 39

If it works, why do I see an error in the error_log. Is there a better way?

[Piranha]

The easiest way is probably to give the result a name. And since this query will always return exactly one row you can change from a while-statement to an if-statement. Another suggestion is to not have the same name for the variable name that holds the result set from mysql and the actual count. It is possible to have the same name, but it may confuse you in some cases. Oh, and use <?php instead of <?. Short tags is removed from PHP6, and may cause problems in versions before that as well.

<?php $result = mysql_query("SELECT COUNT(*) as cou FROM customers WHERE status = '1'");
if ($row = mysql_fetch_array($result)) {
       $count = $row['cou'];
       }
       ?>
<?php echo $count; ?> Active Customers