I know this has been ask many times. But I didn't figure out to make it according to my needs.
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.
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.
But my basic query is :
\App\Model\User::whereNotIn('id', $ids)
               ->where('status', 1)
               ->whereHas('user_location', function($q) use ($lat, $lng, $radius) {
                    /** This is where I'm stuck to write the query **/
             })->select('id', 'firstname')
               ->get();
I don't figure out how to implement the solution in this case.
Thank you in advance for your help
EDIT To be more clear: I need to get the users that are in a 5 kilometers radius.
The distance formula is: √[(x₂ - x₁)² + (y₂ - y₁)²]. This works for any two points in 2D space with coordinates (x₁, y₁) for the first point and (x₂, y₂) for the second point.
php function distance($lat1, $lon1, $lat2, $lon2, $unit) { if (($lat1 == $lat2) && ($lon1 == $lon2)) { return 0; } else { $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 = $ ...
I found the solution, thanks to EddyTheDove and Ohgodwhy.
So this is it:
\App\Model\User::whereNotIn('id', $ids)
           ->where('status', 1)
           ->whereHas('user_location', function($q) use ($radius, $coordinates) { 
                  $q->isWithinMaxDistance($coordinates, $radius);
         })->select('id', 'firstname')
           ->get();
And in my UserLocation Model I have this local scope
public function scopeIsWithinMaxDistance($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]);
}
The original answer by Ohgodwhy is here: Haversine distance calculation between two points in Laravel
EDIT
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 ;
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)
Original article: https://www.plumislandmedia.net/mysql/stored-function-haversine-distance-computation/
Then using eloquent:
\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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