在MySql中使用空间函数查找两个坐标之间的距离的正确方法 [英] Correct way of finding distance between two coordinates using spatial function in MySql

问题描述

我正在尝试使用Mysql和PostgresSQL中的空间函数来计算两个位置之间的距离.我已经从Google取得了经度和纬度.详细信息如下

I am trying to calculate distance between two locations using spatial functions in both Mysql and PostgresSQL. I have taken the latitude and longitude from Google. The details are below

位置一-纬度:42.260223；朗:-71.800010

Location one - Lat: 42.260223; Lon: -71.800010

位置2-纬度:42.245647；朗:-71.802521

Location two - Lat: 42.245647; Lon: -71.802521

使用的SQL查询:

SELECT DISTANCE(GEOMFROMTEXT('Point(42.260223 -71.800010)')，GEOMFROMTEXT('Point(42.245647 -71.802521)'))

SELECT DISTANCE(GEOMFROMTEXT('Point(42.260223 -71.800010)'),GEOMFROMTEXT('Point(42.245647 -71.802521)'))

两个数据库都给出相同的结果0.014790703059697.但是，当我在其他系统中计算距离时，结果却有所不同.请参考以下链接

The both databases are giving the same result 0.014790703059697. But when I calculate distance in other systems the results are different. Please refer the below links

http://www.zip- codes.com/distance_calculator.asp?zip1=01601&zip2=01610&Submit=搜索 = 1.44英里

http://www.distancecheck.com/zipcode- distance.php?start = 01601& end = 01610 = 1.53英里

所以我想知道我的计算方法/查询是否正确.如果错了，那么查询距离的正确方法是什么.

So I want to know whether my calculation method/query is right or not. And if it is wrong, then what is the right way of querying the db for the distance.

推荐答案

简单的答案是使用Haversine公式.假定地球不是一个球体，但这不是一个不好的近似.本演示文稿中对此进行了描述，其中包括许多其他细节:

The simple answer is to use the Haversine formula. This assumes the earth is a sphere, which it isn't, but it's not a bad approximation. This, with lots of other details are described in this presentation:

http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL

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