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我想得到两个邮政编码之间的距离。

HTML表单有一个字段，即邮政编码，英里（距离）。\

具有 7000 条记录的数据库结构。

id | name | zip_code | lat | lng | status|

MySQL 查询

   $query =  "SELECT *, 3963 * acos(cos(radians(90-lat ))*cos(radians(90-'".$result['lat']."'))".
              "+sin(radians(90-lat ))* sin(radians(90-'".$result['lat']."'))".
              "*cos(radians(lng- '".$result['lng']."'))) AS distance FROM table_name".
              " WHERE memberLevel='basic' HAVING (distance < '".$miles."') 
                ORDER BY distance ASC";

我不知道将距离与输入和存储的纬度、经度进行比较的正确 MySQL 查询是什么。

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1 回答 1

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您可以使用空间函数和空间类型列http://dev.mysql.com/doc/refman/5.5/en/spatial-extensions.html

这是使用空间函数的一个很好的例子！ http://www.mysqlperformanceblog.com/2013/10/21/using-the-new-spatial-functions-in-mysql-5-6-for-geo-enabled-applications/

或者你可以定义一个自定义函数

CREATE DEFINER=`test`@`%` FUNCTION `geoDistance`(`lon1` DOUBLE, `lat1` DOUBLE, `lon2` DOUBLE, `lat2` DOUBLE) RETURNS double LANGUAGE SQL DETERMINISTIC NO SQL SQL SECURITY DEFINER COMMENT '' BEGIN DECLARE v DOUBLE; SELECT cos(radians(lat1)) * cos(radians(lat2)) * cos(radians(lon2) - radians(lon1)) + sin(radians(lat1)) * sin(radians(lat2)) INTO v; RETURN IF(v > 1, 0, 6371000 * acos(v)); END

然后打电话

SELECT geoDistance(X(point1), Y(point1), X(spoint2), Y(point2)) result comes in meters

于 2014-05-12T18:53:42.277 回答