PHP MySQL statement - "select as" with "where"

I have the following SQL statement which selects companies from the database (with a stored lat/lng) and displays the nearest 5 locations to the customer's location. This is working perfectly:

$query = sprintf("SELECT company_name, address, telephone, fax,
contact_email, website, url, latitude, longitude, (1.609344 * 3959 *
acos( cos( radians('".$center_lat."') ) * cos( radians( latitude ) ) *

cos( radians( longitude ) - radians('".$center_lng."') ) + sin(
radians('".$center_lat."') ) * sin( radians( latitude ) ) ) ) AS
distance FROM cmsms_module_compdir_companies WHERE status='published'
AND latitude!='' AND longitude!='' ORDER BY distance limit 5 ",
mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat));

However, I'd like to limit the results to only those stores within X distance from the customer's location - say 50 kilometers. I thought I could add the bit in bold below:

$query = sprintf("SELECT company_name, address, telephone, fax,

contact_email, website, url, latitude, longitude, (1.609344 * 3959 *
acos( cos( radians('".$center_lat."') ) * cos( radians( latitude ) ) *
cos( radians( longitude ) - radians('".$center_lng."') ) + sin(
radians('".$center_lat."') ) * sin( radians( latitude ) ) ) ) AS
distance FROM cmsms_module_compdir_companies WHERE status='published'
AND latitude!='' AND longitude!='' AND distance<'50' ORDER BY
distance limit 5 ", mysql_real_escape_string($center_lat),
mysql_real_escape_string($center_lng),
mysql_real_escape_string($center_lat));

...but this returns no results at all.

Any ideas?

Answer

You cannot use calculated value distance in where condition.

plz use having

where ....
having distance < 50

Btw : the unit of distance is kilometer?