通过其他连接获得laravel 5中的纬度和经度

Question

这对我来说非常复杂。我有一个搜索路线 - 效果很好，除了我无法通过邮政编码搜索到最近的lat和一个给定的邮政编码。也就是说，我可以计算出lat和long，但我不确定如何将它集成到我现有的查询中。此查询是没有邮政编码的搜索查询：

$query = DB::table('dogs');
$query->leftJoin('dog_addresses','dogs.id','=','dog_addresses.dog_id');
$query->leftJoin('dog_videos','dogs.id','=','dogs_videos.dog_id');
$query->leftJoin('dogs_breeds','dogs.breed_id','=','dogs_breeds.id');
if($request->input("breed") && $request->input("breed") != "" && $request->input("breed") != "any")
    {
        $breed = Dog_Breed::where("breed_name", $request->input("breed"))->first();
        $query->where('dogs.breed_id', $breed->id);
    }
$results = $query->get();

我有一些东西要添加到查询中以获取邮政编码的纬度和经度：

if($request->input("postcode")) 
{
    $curl = curl_init();
    curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($curl, CURLOPT_URL, "http://api.postcodes.io/postcodes/" . $request->input('postcode'));
    $result = json_decode(curl_exec($curl));
    curl_close($curl);
    $postcode_lat = $result->result->latitude;
    $postcode_long = $result->result->longitude;            
}

这让我得到了我的邮政编码和经度。但是我不知道如何根据dog_addresses表中存在的lat和long列来获取狗的位置，该表连接到dog表。我该怎么做呢？

所以如果我的dog_addresses表有列Lat和Long。

所以狗：

id | user_id | dog_name | age

dog_addresses：

id | dog_id | address_line_1 | town | postcode | lat | long

所以对于我的查询，我需要得到所有的狗，其中繁殖的ID是1，但我想内部加入视频，所以我可以获得所有视频信息和地址信息，但我也想要排序的狗的顺序，他们有多接近是我输入的邮政编码，基于lat和long。

我很困惑。我找到了这个：

( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance

但我仍然不确定如何整合它，或者它对我有什么用处。请帮忙

Answer 1

下面的代码将能够以您想要的方式检索结果。代码说明在线。

if($request->input("postcode")) 
{
    $curl = curl_init();
    curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($curl, CURLOPT_URL, "http://api.postcodes.io/postcodes/" . $request->input('postcode'));
    $result = json_decode(curl_exec($curl));
    curl_close($curl);
    $postcode_lat = $result->result->latitude;
    $postcode_long = $result->result->longitude;   

    $query = DB::table('dogs');
    //Join statement responsible for retrieving dogs addresses based on latitude and longitude in address table.
    $query->join(DB::raw('(SELECT  dog_id, (
                              3959 * acos (
                              cos ( radians($postcode_lat) )
                              * cos( radians( lat ) )
                              * cos( radians( long ) - radians($postcode_long) )
                              + sin ( radians($postcode_lat) )
                              * sin( radians( lat ) )
                            )       
                    )AS distance from dog_addresses) as dog_addresses'), function ($join){
            $join->on('dogs.id', '=', 'dog_addresses.dog_id')

        }); 
    $query->leftJoin('dog_videos','dogs.id','=','dogs_videos.dog_id');
    $query->leftJoin('dogs_breeds','dogs.breed_id','=','dogs_breeds.id');
    if($request->input("breed") && $request->input("breed") != "" && $request->input("breed") != "any")
    {
        $breed = Dog_Breed::where("breed_name", $request->input("breed"))->first();
        $query->where('dogs.breed_id', $breed->id);
    }
    $results = $query->orderBy('dog_addresses.distance', 'ASC') //Ordering the results in ascending order by calculated distance 
                    ->limit(20) //Limiting the results to 20 . Can be changed or removed according to your needs
                    ->get(); //Retrieving the results
}

Answer 2

你可以尝试mysql的GLength函数：

SELECT 
    *,  GLength(LineString(GeomFromText('POINT(latPosCode,lngPosCode)'),GeomFromText('POINT(latInDataBase,lngInDataBase)'),)) AS DISTANCE
FROM dogs           
ORDER BY DISTANCE