Question

我必须根据乘客拉特和日志找到所有20公里的司机。

我有以下两张司机和乘客表。
passenger_id PASSENGER_NAME passenger_lat passenger_log
1 David 123.07 55.03
2 Sunil xyz.pq abc.7

和驱动程序表一样

Driver_id DRIVER_NAME Driver_lat Driver_log
1 Anil 123.07 55.03
2 Jamis xyz.pq abc.7

现在我有了passenger_lat和passenger_log的输入参数。基于我必须返回20公里内的所有驱动程序详细信息

 SELECT *, ( 6371 * acos( cos( radians({$lat}) ) * cos( radians( lat ) ) * cos( radians(   lng ) - radians({$lng}) ) + sin( radians({$lat}) ) * sin( radians( lat ) ) ) ) AS distance FROM positions HAVING distance <= {$radius} ORDER BY distance ASC

plz在php json和mysql中提供答案

Answer 1

<?php

class driver_service
{
    public function find_driver($data)
    {
        $driver_details=array();
        $driver_detail=array();
        $query="select driver_user_id,first_name,last_name,phone_no,cpf_no,driver_email_id,driver_image,current_latitude,current_longitude from driver_user";
        $query_run=  mysql_query($query);
       // $row=  mysql_fetch_assoc($query_run);
        while($row=  mysql_fetch_assoc($query_run))
        {
         $driver=array();
         $dis=  $this->call_distance($data['user_latitude'],$data['user_longitude'],$row['current_latitude'],$row['current_longitude']);


         if($dis<=50)
         {
            $driver['driver_details']=$row; 
            $driver_detail[]=$driver;

         }
      }

      if(empty($driver_detail))
      {
          $driver_details['status']="2~there is no driver between 20 kms";

      }
      else{
      $driver_details['status']="1~success";    
      $driver_details['details']=$driver_detail;
      }
      echo json_encode($driver_details);
    } 
      public function call_distance($lat1, $lon1, $lat2, $lon2) {

      $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 = $dist * 60 * 1.1515;
      return ($miles * 1.609344);
      }


 }
 $driver_service=new driver_service();


if($_REQUEST['action'] == 'find_driver')
{
    $driver_service->find_driver($_REQUEST);

}

Answer 2

class distance
{
    public function all_user()
    {

        $query="select * from driver_user";
        $query_run=  mysql_query($query);
       // $row=  mysql_fetch_assoc($query_run);
        while($row=  mysql_fetch_assoc($query_run))
        {
         $dis=  $this->call_distance(10,20,$row['current_latitude'],$row['current_longitude']);
        }
    }

  public function call_distance($lat1, $lon1, $lat2, $lon2) {

      $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 = $dist * 60 * 1.1515;
      return ($miles * 1.609344);


}
}
$distance=new distance();
$distance->all_user();

如何在纬度和经度的帮助下找到20kms的用户

2 个答案: