Question

我的user类hasMany address类。 address类具有范围查询功能，可获取区域附近的所有地址：

public static function scopeGetByDistance($query,$lat, $lng, int $max_distance)
{
     return $query->selectRaw('id, lat, lng, profileAddress.idMember, ( 3959 * acos( cos( radians( '. $lat.') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(' . $lng . ') ) + sin( radians(' . $lat .') ) * sin( radians(lat) ) ) ) AS distance')
                  ->having('distance', '<', $max_distance );
}

我想让所有地址为max_distance到(lat,lng)点的用户，并希望按照该距离对其进行排序。

如何使用查询生成器来实现？我尝试过：

  $users = \App\User::leftJoin('addresses', function($join){
    $join->getByDistance(44.81, 20.46, 20);
  })
    ->select(DB::raw('MIN(distance) as closest'))
    ->groubBy('user.id')
    ->orderBy('closest')
    ->get();

但这会导致

方法Illuminate \ Database \ Query \ JoinClause :: getByDistance不存在。

Answer 1

尝试这样的事情

   $users = DB::table('users as u')
   ->join('address as ad','u.id','=','ad.uset_id');
   ->select(DB::raw('(6371 * acos(cos(radians('.$lat.')) * cos(radians(lat)) * cos(radians(lng) - radians('.$long.')) + sin(radians('.$lat.')) * sin(radians(lat )))) AS distance'))
   ->having('distance','<',$max_distance)         
   ->get();

Answer 2

好吧，我刚刚发现join and relationships are not supported by Laravel

这是我最后所做的：

 $users = \App\User::leftJoin('profileAddress','users.id','=','profileAddress.id')
   ->selectRaw('profileAddress.*,   MIN(3959 * acos( cos( radians( '. $lat.') ) * cos( radians( profileAddress.lat ) ) * cos( radians( profileAddress.lng ) - radians(' . $lng . ') ) + sin( radians(' . $lat .') ) *  sin( radians(lat) ) ) ) as closest')
    ->groupBy('users.id')
    ->having('closest', '<', 20)
    ->orderBy('closest')
    ->get();

Laravel：使用联接和范围查询

2 个答案: