使用PHP和MySQL的地理位置半徑搜索

Question

我正在開發一個求職門戶，最終用戶將根據他們的位置和鄰近半徑搜索作業。目前REST API由SLIM和MySQL提供支持。 在Jobs表中，每行都有經度和緯度。我在這裏搜索了這個，偶然發現了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 

現在我還有兩個問題。

1. 如果只有MySQL查詢足夠了，或者我必須使用它設置Elastic Search？
2. 如果有任何庫或軟件包可以幫助我解決這個特定問題？

而且如果您可以爲搜索性能提供任何可能有用的建議。

Answer 1

我已經下面 -

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; 

我做標杆無線發現這個數據比穆克什的答案稍快，比@ jision的答案要好2倍。

Answer 2

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; 

凡

$ LAT =用戶緯度

$長=用戶經度

Answer 3

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; 
       } 
     }