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我的数据库中有一个表，它存储来自谷歌地图的位置以及它们返回的坐标。以前我将 lat 和 lng 存储为小数点，在我将 MySQL 版本升级到 8 后，发现有预定义的数据类型和函数来处理坐标和距离计算。我创建了如下表格，并从谷歌地图中插入了一些随机值。

CREATE TABLE `Location` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `lat` varchar(50) NOT NULL,
  `lng` varchar(50) NOT NULL,
  `coordinates` point NOT NULL,
  `name` varchar(500) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uk_lat_lng` (`lat`,`lng`),
  SPATIAL KEY `sk_coordinates` (`coordinates`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

插入的样本数据

INSERT INTO Location (name, lat, lng, coordinates)
SELECT 'Thayyeni Koomban', '12.33984592973732', '75.41285991668701', POINT(12.33984592973732, 75.41285991668701) UNION
SELECT 'Tejaswini river rafting', '12.29208143873455', '75.4130744934082', POINT(12.29208143873455, 75.4130744934082) UNION
SELECT 'Mossy Forest, Cameron Highlands', '4.5242225', '101.38192709999998', POINT(4.5242225, 101.38192709999998) UNION
SELECT 'Dhanushkodi Point', '9.152226599999999', '79.44291569999996', POINT(9.152226599999999, 79.44291569999996) UNION
SELECT 'KL Sentral', '3.13333', '101.68667000000005', POINT(3.13333, 101.68667000000005) UNION
SELECT 'Kozhippara Waterfalls', '11.3537295', '76.10803290000001', POINT(11.3537295, 76.10803290000001) UNION
SELECT 'Laguna Honda Hospital and Rehabilitation Center', '37.7492806', '-122.45702240000003', POINT(37.7492806, -122.45702240000003) UNION
SELECT 'Singapore Zoo', '1.4043485', '103.79302299999995', POINT(1.4043485, 103.79302299999995) UNION
SELECT 'Taj Mahal', '27.1750151', '78.04215520000002', POINT(27.1750151, 78.04215520000002) UNION
SELECT 'Sea View Point', '11.2643567', '75.76153939999995', POINT(11.2643567, 75.76153939999995) UNION
SELECT 'Club Mahindra Ashtamudi', '8.965749299999997', '76.57136119999996', POINT(8.965749299999997, 76.57136119999996) UNION
SELECT 'Kollam Beach', '8.8756778', '76.58891629999994', POINT(8.8756778, 76.58891629999994) UNION
SELECT 'Tropical Islands', '52.03892399999999', '13.748616999999967', POINT(52.03892399999999, 13.748616999999967) UNION
SELECT 'Taroko National Park', '24.15870679999999', '121.62162969999997', POINT(24.15870679999999, 121.62162969999997) UNION
SELECT 'Lake Tyrrell', '-35.2553505', '142.8419824', POINT(-35.2553505, 142.8419824) UNION
SELECT 'Karthika Regency', '10.0192948', '76.30539509999994', POINT(10.0192948, 76.30539509999994) UNION
SELECT 'Sentosa', '1.2494041', '103.83032090000006', POINT(1.2494041, 103.83032090000006) UNION
SELECT 'Kovalam Beach', '12.7902597', '80.25390390000007', POINT(12.7902597, 80.25390390000007) UNION
SELECT 'Torres del Paine National Park', '-50.9423262', '-73.40678789999998', POINT(-50.9423262, -73.40678789999998) UNION
SELECT 'Niagara Falls', '43.0828162', '-79.07416289999998', POINT(43.0828162, -79.07416289999998);

了解 google 坐标使用 3857 的 SRID，因此使用 lat 和 lng 的不同组合创建了以下查询来生成点。但似乎没有一个距离是准确的。

SELECT 
    name, 
    lat, 
    lng, 
    ST_Distance(ST_SRID(coordinates, 3857), ST_GeomFromText('POINT(10.0120262 76.3586236)', 3857)) AS distance,
    ST_Distance(ST_SRID(POINT(lat, lng), 3857), ST_GeomFromText('POINT(10.0120262 76.3586236)', 3857)) AS lat_lng,
    ST_Distance(ST_SRID(POINT(lng, lat), 3857), ST_GeomFromText('POINT(10.0120262 76.3586236)', 3857)) AS lng_lat_first,
    ST_Distance(ST_SRID(POINT(lng, lat), 3857), ST_GeomFromText('POINT(76.3586236 10.0120262)', 3857)) AS lng_lat_all,
    ST_Distance(ST_SRID(POINT(lng, lat), 3857), ST_SRID(POINT(76.3586236, 10.0120262), 3857)) AS lng_lat_all_x,
    ST_Distance(ST_SRID(POINT(lng, lat), 3857), ST_SRID(POINT(10.0120262, 76.3586236), 3857)) AS lng_lat_all_y, 
    ST_Distance(ST_SRID(POINT(lat, lng), 3857), ST_SRID(POINT(10.0120262, 76.3586236), 3857)) AS lng_lat_all_z   
from test.Location;

结果

我是以正确的方式做这件事还是遗漏了什么？提前致谢。

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1 回答 1

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我不确定。但是 ST_Distance 采用两个参数。这里是三个参数。示例是

mysql> SET @g1 = Point(1,1);
mysql> SET @g2 = Point(2,2);
mysql> SELECT ST_Distance(@g1, @g2); .

于 2018-11-16T13:07:42.937 回答