[SOLVED] FLOOR in mysql statement

[aldrin151]

Hi everyone...I've been trying to round of by FLOOR in mysql statement the value of AS distance below...does anyone know how to write the correct way...? Reason is that when I send a param of miles in a round number format I want it to match this value.

 

  $sSQL = "SELECT *, (ACOS((SIN(RADIANS(" .$lat."))*SIN(RADIANS(latitude))) + (COS(RADIANS(" .$lat."))*COS(RADIANS(latitude))*COS(RADIANS(longitude)-RADIANS(" .$lng. ")))) * ".$unit1.") AS distance FROM users WHERE (ACOS((SIN(RADIANS(" .$lat. "))*SIN(RADIANS(latitude))) + (COS(RADIANS(" .$lat. "))*COS(RADIANS(latitude))*COS(RADIANS(longitude)-RADIANS(" .$lng. ")))) * ".$unit1.") <= " .$miles. "";

 

Thanks

[aldrin151]

Funny I solved my own problem....this may help someone out there

 

 

$sSQL = "SELECT *, FLOOR((ACOS((SIN(RADIANS(" .$lat."))*SIN(RADIANS(latitude))) + (COS(RADIANS(" .$lat."))*COS(RADIANS(latitude))*COS(RADIANS(longitude)-RADIANS(" .$lng. ")))) * ".$unit1.")) AS distance FROM users WHERE FLOOR((ACOS((SIN(RADIANS(" .$lat. "))*SIN(RADIANS(latitude))) + (COS(RADIANS(" .$lat. "))*COS(RADIANS(latitude))*COS(RADIANS(longitude)-RADIANS(" .$lng. ")))) * ".$unit1.")) <= " .$miles. "";

  $sOrder = "";