Question

我试图编写一个搜索位置附近的搜索栏。现在，它是硬编码的。该语句有效，直到我为经度和纬度添加like语句。有更好的方法吗？

$searchString = explode(' ', $_GET["searchString"]);
$useLocation = $_GET["useCurrentLocation"]; 
if($useLocation == 1){
    foreach($searchString AS $key){
        $stmt = $db->prepare("SELECT * FROM locations WHERE Name LIKE '%{$key}%' AND Longitude Like '-79.8579' And Latitude Like '43.5237'"); 
        $stmt->execute();
        $result = $stmt->get_result();
        while($row = $result->fetch_assoc()){
            echo $row['Name'];
        }

    }
}

Answer 1

您可以在互联网上找到许多关于“如何按位置搜索”的解决方案

Stackoverflow上有一些答案：

或者在Google上：

http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

这是我对此的最后一个问题：

我通过Google Maps API获得经度/纬度。输入地址，并获取GPS坐标。之后，用SQL计算位置和周围的对象

$ longitude_ref = 7.694731; $ latitude_ref = 48.560719;

SELECT *
  FROM maps
  WHERE (
    6366 *
    acos(
      cos(radians(48.560719)) -- $latitude_ref
      *
      cos(radians(`latitude`))
      *
      cos(radians(`longitude`) - radians($longitude_ref)) -- $longitude_ref
      +
      sin(radians(48.560719)) * sin(radians(`latitude`)) -- $latitude_ref
    )
  ) <= 2 -- r in Km (here: 2 km)

您还可以计算两点之间的距离：

SELECT pg.id, pg.*, pg1.*, 6366 * 2 *
    atan2(
      sqrt(
        SIN((RADIANS(pg1.latitude) - RADIANS(pg.latitude)) / 2) * SIN((RADIANS(pg1.latitude) - RADIANS(pg.latitude)) / 2) + COS(RADIANS(pg.latitude)) * COS(RADIANS(pg1.latitude)) * SIN((RADIANS(pg1.longitude) - RADIANS(pg.longitude)) / 2) * SIN((RADIANS(pg1.longitude) - RADIANS(pg.longitude)) / 2)    
      ),
      sqrt(
        1 - SIN((RADIANS(pg1.latitude) - RADIANS(pg.latitude)) / 2) * SIN((RADIANS(pg1.latitude) - RADIANS(pg.latitude)) / 2) + COS(RADIANS(pg.latitude)) * COS(RADIANS(pg1.latitude)) * SIN((RADIANS(pg1.longitude) - RADIANS(pg.longitude)) / 2) * SIN((RADIANS(pg1.longitude) - RADIANS(pg.longitude)) / 2)
      )
    ) as calculated_distance_final_km
  FROM perso_gps pg
  WHERE pg.calculated_dist IS NULL

互联网上的许多例子：）

SQL搜索＆＃34;附近＆＃34;经度和纬度

1 个答案: