MySQL之间的查询2 lat，long给出不同于PHP代码的数字

Question

所以我有这个PHP代码：

$lat1 = 39.210637;
$lng1 = -105.534551;
$lat2 = 39.725865;
$lng2 = -105.016820;

$theta = $lng1 - $lng2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);

$miles = $dist * 60 * 1.1515;

$kilometers = $miles * 1.60934;

$meters = $kilometers * 1000;

返回：72503米

我尝试在MySQL中编写完全相同的代码：

ROUND((RADIANS(ACOS(SIN(RADIANS(39.210637)) * SIN(RADIANS(39.725865)) + COS(RADIANS(39.210637)) * COS(RADIANS(39.725865)) * COS(RADIANS(-105.534551 - (-105.016820)))))) * 60 * 1.1515 * 1.60934 * 1000)

返回：22米

为什么他们应该是相同的功能呢？我做错了吗？

谢谢！

Answer 1

糟糕，看起来我错过了一个应该是MySQL中的DEGREES（）的rad2deg，所以代码应该是：

ROUND((DEGREES(ACOS(SIN(RADIANS(39.210637)) * SIN(RADIANS(39.725865)) + COS(RADIANS(39.210637)) * COS(RADIANS(39.725865)) * COS(RADIANS(-105.534551 + 105.016820))))) * 60 * 1.1515 * 1.60934 * 1000)

现在它运作正常。