Got the query, Cant figure out the right code to display it..

[SBuzz]

Hi,

the following query:

PHP Code: [Select]

$query1 = "SELECT d.value, c.value, d.user_id, (SQRT(POW((b.x - a.x), 2) + POW((b.y - a.y), 2))/1000) * 0.621 AS distance
 FROM postcodes a, postcodes b, wp_bp_xprofile_data c JOIN  wp_bp_xprofile_data d 
     ON c.user_id=d.user_id AND d.field_id=1 
 WHERE a.outcode = '".$postcode."' AND b.outcode = c.value
 HAVING (distance < '".$area."')
ORDER BY distance asc ";


when i run it on the database, returns exactly what i want. 

I just need the php code to display all the info.

My code (trunkated):

PHP Code: [Select]

// display results

while ($list = mysql_fetch_array($result1)) {
    $user_postcode = $list['value'];

$distance = $list['distance'];

$distance = round($distance);
echo "<td colspan=\"3\" align=\"left\"><strong>$user_postcode</strong></td>";


Will display only one of the 'value', from the wp_bp_xprofile_data table..

Any help? Any more info you need?

Thanks

[SBuzz]

In short,  how to define the *other* returned 'value' to a $variable?

[SBuzz]

I got it..

Define them 1st using AS, then getting them back with

mysql_fetch_assoc.

Thanks for the help. =)