我有不同城市的不同酒店的数据库。每个旅馆都有纬度和经度。
我是否可以通过SQL获取周围的酒店,比如25公里以内。我正在使用mySQL。
发布于 2011-10-26 14:22:41
首先,在控制台模式下运行此命令。它将创建一个新的MySQL函数(请参见下面的phpMyAdmin)。
DELIMITER |
DROP FUNCTION IF EXISTS GPS_Distance_Meters|
CREATE FUNCTION GPS_Distance_Meters (lat1 DOUBLE, lng1 DOUBLE, lat2 DOUBLE, lng2 DOUBLE) RETURNS DOUBLE
BEGIN
DECLARE rlo1 DOUBLE;
DECLARE rla1 DOUBLE;
DECLARE rlo2 DOUBLE;
DECLARE rla2 DOUBLE;
DECLARE dlo DOUBLE;
DECLARE dla DOUBLE;
DECLARE a DOUBLE;
SET rlo1 = RADIANS(lng1);
SET rla1 = RADIANS(lat1);
SET rlo2 = RADIANS(lng2);
SET rla2 = RADIANS(lat2);
SET dlo = (rlo2 - rlo1) / 2;
SET dla = (rla2 - rla1) / 2;
SET a = SIN(dla) * SIN(dla) + COS(rla1) * COS(rla2) * SIN(dlo) * SIN(dlo);
RETURN (6378137 * 2 * ATAN2(SQRT(a), SQRT(1 - a)));
END|
DELIMITER ;也可以在phpMyAdmin中运行,但记住将分隔符更改为|,而不是;
DROP FUNCTION IF EXISTS GPS_Distance_Meters|
CREATE FUNCTION GPS_Distance_Meters (lat1 DOUBLE, lng1 DOUBLE, lat2 DOUBLE, lng2 DOUBLE) RETURNS DOUBLE
BEGIN
DECLARE rlo1 DOUBLE;
DECLARE rla1 DOUBLE;
DECLARE rlo2 DOUBLE;
DECLARE rla2 DOUBLE;
DECLARE dlo DOUBLE;
DECLARE dla DOUBLE;
DECLARE a DOUBLE;
SET rlo1 = RADIANS(lng1);
SET rla1 = RADIANS(lat1);
SET rlo2 = RADIANS(lng2);
SET rla2 = RADIANS(lat2);
SET dlo = (rlo2 - rlo1) / 2;
SET dla = (rla2 - rla1) / 2;
SET a = SIN(dla) * SIN(dla) + COS(rla1) * COS(rla2) * SIN(dlo) * SIN(dlo);
RETURN (6378137 * 2 * ATAN2(SQRT(a), SQRT(1 - a)));
END|现在使用函数。示例:
SELECT *
FROM restaurants
WHERE GPS_Distance_Meters('48.858205', '2.294359', lat, lng) < 1000 ORDER BY proximite ASC
LIMIT 10在这里,我们从48.858205,2.294359 (与野外lat和lng相比) 1000米(1 1KM)范围内搜索limit 10结果。
https://stackoverflow.com/questions/7898964
复制相似问题