为什么我无法使用PHP / MySQL和纬度/经度从此查询中获取数组？

Question

I have two tables but for the purposes of this question, I will only use one. I am using the latitude and longitude for a city to find airports within 50 miles. As far as I know, it's working but I can't seem to put this into an array with mysql_fetch_array... It's something stupid or something small... it always seems to be one or the other. After trying several things, this is where I am at right now with the code:

<?php 
  require('dbconnect.php');
  //airports Table Columns
    //iata_code
    //airport_name
    //airport_name_clean
    //city_id
    //airport_lat
    //airport_long

$cityLat = "25.788969"; //Miami
$cityLong = "-80.226439"; //Miami
$distance = "10"; //miles?

$airportQuery = mysql_query("select airport_name, 
   ( 3959 * acos( cos( radians($cityLat) ) 
          * cos( radians( locations.lat ) ) 
          * cos( radians( locations.lng ) - radians($cityLong) ) 
          + sin( radians($cityLat) ) 
          * sin( radians( locations.lat ) ) ) ) AS distance 
from airports 
and locations.lat between X1 and X2 
and locations.Long between y1 and y2
having distance < $distance ORDER BY distance;
");

while($airports = mysql_fetch_array($airportQuery))
    {
        echo $airports['airport_name'] . "<br />";
    }

?>

As always, any help will be greatly appreciated. Thank you so much for your help!

Answer 1

Your query looks suspicious, I think you have an and when you should have a where :

$airportQuery = mysql_query("select airport_name, 
   ( 3959 * acos( cos( radians($cityLat) ) 
          * cos( radians( locations.lat ) ) 
          * cos( radians( locations.lng ) - radians($cityLong) ) 
          + sin( radians($cityLat) ) 
          * sin( radians( locations.lat ) ) ) ) AS distance 
from airports 
WHERE locations.lat between X1 and X2 
and locations.Long between y1 and y2
having distance < $distance ORDER BY distance;
");

You should check the return value for errors, that might be informative as well.