在SQL查询中查找两个纬度和经度之间的距离

Question

我有一个名为＆＃34的承包商＆＃34;并且该表有5个字段

现在，我已经选择了另一个特定记录，而且我们选择了这些记录，它们是：＆/ p>

纬度：19.2494730，经度：72.8612790

我的SQL查询是：

SELECT *, round( (3959 * acos( cos( radians(19.2494730) ) * cos( radians( tbl.latitude ) ) * cos( radians( tbl.longitude ) - radians(72.8612790) ) + sin( radians(19.2494730) ) * sin( radians( tbl.latitude ) ) ) ),2) AS distance
FROM `contractors` AS tbl

并给出以下结果：

但距离（它在Miles中我认为）是不正确的，因为当我运行以下JavaScript代码时，它会给我一些准确的结果。

function distance(lat1, lon1, lat2, lon2, unit) {
    var radlat1 = Math.PI * lat1/180
    var radlat2 = Math.PI * lat2/180
    var radlon1 = Math.PI * lon1/180
    var radlon2 = Math.PI * lon2/180
    var theta = lon1-lon2
    var radtheta = Math.PI * theta/180
    var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
    dist = Math.acos(dist)
    dist = dist * 180/Math.PI
    dist = dist * 60 * 1.1515
    if (unit=="K") { dist = dist * 1.609344 }
    if (unit=="N") { dist = dist * 0.8684 }
    alert(dist)
}            
distance(19.2494730, 72.8612790, 19.281085, 72.855994, 'K');

我还有一个PHP代码段，可以提供更准确的结果

function distance($lat1, $lon1, $lat2, $lon2, $unit) {

  $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 = $dist * 60 * 1.1515;
  $unit = strtoupper($unit);

  if ($unit == "K") {
    return ($miles * 1.609344);
  } else if ($unit == "N") {
      return ($miles * 0.8684);
    } else {
        return $miles;
      }
}

echo distance(19.2494730, 72.8612790,19.281085, 72.855994, "M") . " Miles<br>";

任何人都可以帮助使上述SQL查询正确，这样我就能给出所有区域和特定Lat，Long之间的准确直线距离。

Answer 1

Haversine Formula？

CREATE function [dbo].[GetDistance](
    @lat1  decimal(18,10),
    @lon1  decimal(18,10),
    @lat2  decimal(18,10),
    @lon2  decimal(18,10)
    )
    returns decimal(18,4)
    as
    begin
        declare @radius decimal(18,10)

        declare @a decimal(18,10)
        declare @distance decimal(18,4)

        -- Sets average radius of Earth in Kilometers
        set @radius = 6371.0E

        -- Convert degrees to radians
        set @lon1 = radians( @lon1)
        set @lon2 = radians( @lon2)
        set @lat1 = radians( @lat1 )
        set @lat2 = radians( @lat2 )

        set @a = sqrt(square(sin((@lat2-@lat1)/2.0E)) + 
            (cos(@lat1) * cos(@lat2) * square(sin((@lon2-@lon1)/2.0E))) )

        set @distance =
            @radius * ( 2.0E *asin(case when 1.0E < @a then 1.0E else @a end ))

        return @distance
    end