使用 PHP 和 MySQL 进行地理位置半径搜索

Question

我正在开发一个工作门户，最终用户将在其中根据他们的位置和邻近半径搜索工作。目前，REST API 由 SLIM 和 MySQL 提供支持。 在 Jobs 表中，每一行都有纬度和经度。我在这里搜索过这个并偶然发现了Geo Distance search with MySQL。

SELECT latitude, longitude, SQRT(
POW(69.1 * (latitude - $lat), 2) +
POW(69.1 * ($long - $long) * COS(latitude / 57.3), 2)) AS distance
FROM TableName HAVING distance < 25 ORDER BY distance;


Where 

$lat =  User latitude

$long = User longitude

现在我还有另外两个问题。

1. 如果只有 MySQL 查询就足够了，还是我必须使用它设置 Elastic Search？
2. 是否有任何库或包可以帮助我解决这个特定问题？

如果您能对搜索性能提出任何有用的建议。

Answer 1

SELECT latitude, longitude, SQRT(
POW(69.1 * (latitude - $lat), 2) +
POW(69.1 * ($long - longitude) * COS(latitude / 57.3), 2)) AS distance
FROM TableName HAVING distance < 25 ORDER BY distance;

在哪里

$lat = 用户纬度

$long = 用户经度

Answer 2

You have two option to find job by location 

you required 
1. Search users lat long
2. Job users Posted Lat long (it may be job location or job posted user location from database)
3. Query will be like - 
SELECT zip,Round(((ACOS(SIN('$lat' * PI() / 180) * SIN(latitude * PI() / 180) + COS('$lat' * PI() / 180) * COS(latitude * PI() / 180) * COS(('$lon'-longitude) * PI() / 180)) * 180 / PI()) * 60 * 1.1515),(2)) AS distance FROM Jobs Having distance <= 30 

Else 
1. You have to call all jobs data in single query.

     foreach($joblist as $job){
        $milesresult = $this->calculateDistance($user_lat,$user_lon,$job['latitude'],$jobr['longitude']);
        $miles = explode("-",$vendor['miles']);
         $vendor_max_miles = $miles[1];
        }

2. PHP function for lat long

      function calculateDistance($lat1, $lon1, $lat2, $lon2, $unit) {

            $theta = $lon1 - $lon2;

            $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));

            $dist = acos($dist);

             $dist = rad2deg($dist);

              $miles = $dist * 60 * 1.1515;
            $unit = strtoupper($unit);

            if ($unit == "K") {
              return ($miles * 1.609344);
            } else if ($unit == "N") {
                return ($miles * 0.8684);
              } else {
                  return $miles;
                }
        }

Answer 3

我已经和下面一起去了-

SELECT id,
       name,
       lat,
       lng,
       ROUND((6371 * acos(
                       cos(radians($lat)) * cos(radians(lat)) * cos(radians(lng) - radians($lng)) +
                       sin(radians($lat)) * sin(radians(lat)))), (2)) AS distance
FROM jobs
HAVING distance < 50
ORDER BY distance;

我已经对数据进行了基准测试，发现这比 Mukesh 的答案略快，比 @jision 的答案好 2 倍。