php 使用纬度和经度的 40 公里半径
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/8850336/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Radius of 40 kilometers using latitude and longitude
提问by DerVO
Possible Duplicate:
Radius of multiple latitude/longitude points
distance calculations in mysql queries
select within 20 kilometers based on latitude/longitude
I have a MyISAM table, which the data is stored in. The data includes latitude and longitude for Google Maps. What I am trying to achieve is to select all objects within the radius of 40 kilometers of the sample latitude and longitude.
我有一个 MyISAM 表，数据存储在其中。数据包括谷歌地图的纬度和经度。我想要实现的是选择样本经纬度40公里范围内的所有对象。
Thanks!
谢谢！
回答by DerVO
I use this query to get all points within a $radiusaround $lat/$lng:
我使用此查询来获取$radius周围$lat/内的所有点$lng：
SELECT
*,
( 6371 * acos( cos( radians({$lat}) ) * cos( radians( `lat` ) ) * cos( radians( `lng` ) - radians({$lng}) ) + sin( radians({$lat}) ) * sin( radians( `lat` ) ) ) ) AS distance
FROM `positions`
HAVING distance <= {$radius}
ORDER BY distance ASC
6371 is the earth radius im km. And I have not invented it:
http://code.google.com/intl/en/apis/maps/articles/phpsqlsearch.htmlhttps://developers.google.com/maps/articles/phpsqlsearch_v3#findnearsql(edit: URL changed)
6371 是地球半径 im km。我还没有发明它：http:
//code.google.com/intl/en/apis/maps/articles/phpsqlsearch.htmlhttps://developers.google.com/maps/articles/phpsqlsearch_v3#findnearsql（编辑：网址已更改）
