如何用口才来查询两个坐标之间的距离 [英] How to query distances between two coordinates with Eloquent

问题描述

我知道这个问题已经问了很多遍了.但是我没有想到要根据自己的需要来做.

I know this has been ask many times. But I didn't figure out to make it according to my needs.

我需要从另一个用户那里查询最近的用户.基本上，我有一个 users 表，该表与 users_locations 表具有一对一关系，该表具有纬度和经度字段.

I need to query the nearest users from another user. Basically, I have a users table this table has a one to one relation with the users_locations table which has a latitude and a longitude field.

所以我看过了 https://laravel.io/forum/04-23-2014-convert-this-geolocation-query-to-query-builder-for?page=1 这可能是最好的解决方案.

So I've seen this https://laravel.io/forum/04-23-2014-convert-this-geolocation-query-to-query-builder-for?page=1 and this may be the best solution.

但是我的基本查询是:

\App\Model\User::whereNotIn('id', $ids)
               ->where('status', 1)
               ->whereHas('user_location', function($q) use ($lat, $lng, $radius) {
                    /** This is where I'm stuck to write the query **/
             })->select('id', 'firstname')
               ->get();

在这种情况下，我不知道如何实施该解决方案.

I don't figure out how to implement the solution in this case.

预先感谢您的帮助

编辑更清楚地说:我需要获得半径为5公里的用户.

EDIT To be more clear: I need to get the users that are in a 5 kilometers radius.

推荐答案

感谢EddyTheDove和Ohgodwhy，我找到了解决方案.

I found the solution, thanks to EddyTheDove and Ohgodwhy.

就这样:

\App\Model\User::whereNotIn('id', $ids)
           ->where('status', 1)
           ->whereHas('user_location', function($q) use ($radius, $coordinates) { 
                  $q->isWithinMaxDistance($coordinates, $radius);
         })->select('id', 'firstname')
           ->get();

在我的 UserLocation 模型中，我具有此本地范围

And in my UserLocation Model I have this local scope

public function scopeIsWithinMaxDistance($query, $coordinates, $radius = 5) {

    $haversine = "(6371 * acos(cos(radians(" . $coordinates['latitude'] . ")) 
                    * cos(radians(`latitude`)) 
                    * cos(radians(`longitude`) 
                    - radians(" . $coordinates['longitude'] . ")) 
                    + sin(radians(" . $coordinates['latitude'] . ")) 
                    * sin(radians(`latitude`))))";

    return $query->select('id', 'users_id', 'cities_id')
                 ->selectRaw("{$haversine} AS distance")
                 ->whereRaw("{$haversine} < ?", [$radius]);
}

Ohgodwhy的原始答案在这里:在Laravel中两点之间的Haversine距离计算

The original answer by Ohgodwhy is here: Haversine distance calculation between two points in Laravel

编辑

使用MySQL中存储的函数执行此操作的另一种方法:

Another way to perform it with stored functions in MySQL:

    DELIMITER $$
DROP FUNCTION IF EXISTS haversine$$

CREATE FUNCTION haversine(
        lat1 FLOAT, lon1 FLOAT,
        lat2 FLOAT, lon2 FLOAT
     ) RETURNS FLOAT
    NO SQL DETERMINISTIC
    COMMENT 'Returns the distance in degrees on the Earth
             between two known points of latitude and longitude'
BEGIN
    RETURN DEGREES(ACOS(
              COS(RADIANS(lat1)) *
              COS(RADIANS(lat2)) *
              COS(RADIANS(lon2) - RADIANS(lon1)) +
              SIN(RADIANS(lat1)) * SIN(RADIANS(lat2))
            ));
END$$

DELIMITER ;

我乘以111.045，将结果转换为km.(我不确定该值是否正确，我发现许多其他值与该值相距不远，因此，如果有人对此有精确度，可能会很好)

I multiply by 111.045 to convert the result in km. (I'm not sure that this value is right, I found many others values not far from this one so if someone have precision about it, it could be nice)

原始文章: https://www.plumislandmedia.net/mysql/stored-function-haversine-distance-computation/

然后用雄辩的话:

\App\Model\User::whereNotIn('id', $ids)
       ->where('status', 1)
       ->whereHas('user_location', function($q) use ($radius, $coordinates) { 
            $q->whereRaw("111.045*haversine(latitude, longitude, '{$coordinates['latitude']}', '{$coordinates['longitude']}') <= " . $radius]);
     })->select('id', 'firstname')
       ->get();

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