Question

我有两张桌子但是出于这个问题的目的，我只会使用一张桌子。我正在使用纬度和经度来查找50英里范围内的城市。据我所知，它正在工作，但我似乎无法将它放入一个带有mysql_fetch_array的数组中......这是一些愚蠢或小的东西......它似乎总是一个或另一个。在尝试了几件事后，这就是我现在所处的代码：

<?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 />";
    }

?>

与往常一样，任何帮助将不胜感激。非常感谢你的帮助！

Answer 1

您的查询看起来很可疑，我认为您应该and时有一个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;
");

您应该检查错误的返回值，这也可能是提供信息的。

为什么我无法使用PHP / MySQL和纬度/经度从此查询中获取数组？

1 个答案: