Question

我有一个MySQL例程，当通过纬度和经度使用Haversin方程时，在半径50英里范围内获得记录。

虽然这很好用，并且非常快速（考虑到它正在搜索82k记录），但我认为通过使用POINT列创建类似的过程可以获得更好的性能。

所以，在我的表中，我创建了一个名为Location的额外列，给它一个POINT数据类型，更新了我的数据以传递lat＆amp; lon到Location列。数据有效，并且很好。并添加了Spatial Index

问题是，如何转换以下查询以使用Location列，而不是lat和lon列。

SET @LAT := '37.953';
SET @LON := '-105.688';

SELECT DISTINCT
BPZ.`store_id`,         
3956 * 2 * ASIN(SQRT(POWER(SIN((@LAT - abs(Z.`lat`)) * pi()/180 / 2),2) + COS(@LAT * pi()/180 ) * COS(abs(Z.`lat`) *  pi()/180) * POWER(SIN((@LON - Z.`lon`) *  pi()/180 / 2), 2))) as distance,
c.`name`,c.`address`,c.`city`,c.`state`,c.`phone`,c.`zip`,c.`premise_type`
FROM
`zip_codes` as Z, 
`brand_product_zip` as BPZ
LEFT JOIN `customers` c ON c.`store_id` = BPZ.`store_id`
WHERE
BPZ.`zip` = Z.`zip`
AND 
3956 * 2 * ASIN(SQRT(POWER(SIN((@LAT - abs(Z.`lat`)) * pi()/180 / 2),2) + COS(@LAT * pi()/180 ) * COS(abs(Z.`lat`) *  pi()/180) * POWER(SIN((@LON - Z.`lon`) *  pi()/180 / 2), 2))) <= 50
ORDER BY
distance LIMIT 20

我知道之前已经提到过，但是，我看到的所有内容都指向基于lat和lon而不是POINT列的计算

更新代码：

SET @lat = 41.92;
SET @lon = -72.65;
SET @kmRange = 80.4672; -- = 50 Miles

SELECT *, (3956 * 2 * ASIN(SQRT(POWER(SIN((@lat - abs(`lat`)) * pi()/180 / 2),2) + COS(@lat * pi()/180 ) * COS(abs(`lat`) *  pi()/180) * POWER(SIN((lon - `lon`) *  pi()/180 / 2), 2)))) as distance
FROM    `zip_codes`
WHERE   MBRContains(LineString(Point(@lat + @kmRange / 111.1, @lon + @kmRange / (111.1 / COS(RADIANS(@lat)))), Point(@lat - @kmRange / 111.1, @lon - @kmRange / (111.1 / COS(RADIANS(@lat))))), `Location`)
Order By distance
LIMIT 20

Answer 1

您是否研究过希尔伯特曲线解决方案？空间索引无法提供准确的解决方案？。使用mysql空间索引，您可以使用mbrcontains：

CREATE TABLE lastcrawl (id INT NOT NULL PRIMARY KEY, pnt POINT NOT NULL) ENGINE=MyISAM;

INSERT
INTO    lastcrawl
VALUES  (1, POINT(40, -100));

SET @lat = 40;
SET @lon = -100;

SELECT  *
FROM    lastcrawl
WHERE   MBRContains
                (
                LineString
                        (
                        Point
                                 (
                                 @lat + 10 / 111.1,
                                 @lon + 10 / ( 111.1 / COS(RADIANS(@lat)))
                                 ),
                        Point    (
                                 @lat - 10 / 111.1,
                                 @lon - 10 / ( 111.1 / COS(RADIANS(@lat)))
                                 )
                        ),
                pnt
                );

请看这里：MySQL - selecting near a spatial point。这里：http://www.drdobbs.com/database/space-filling-curves-in-geospatial-appli/184410998

Answer 2

文章Nearest-location finder for MySQL详细解释了各种选项，以及与Spatial Extensions starting with MySQL 5.6一起使用的最佳选择。

从文章中，此示例查询列出了距离给定坐标（42.81，-70.81）半径50英里范围内的邮政编码：

SELECT zip, primary_city,
       latitude, longitude, distance_in_mi
  FROM (
SELECT zip, primary_city, latitude, longitude,r,
       69.0 * DEGREES(ACOS(COS(RADIANS(latpoint))
                 * COS(RADIANS(latitude))
                 * COS(RADIANS(longpoint) - RADIANS(longitude))
                 + SIN(RADIANS(latpoint))
                 * SIN(RADIANS(latitude)))) AS distance_in_mi
 FROM zip
 JOIN (
        SELECT  42.81  AS latpoint,  -70.81 AS longpoint, 50.0 AS r
   ) AS p
 WHERE latitude
  BETWEEN latpoint  - (r / 69)
      AND latpoint  + (r / 69)
   AND longitude
  BETWEEN longpoint - (r / (69 * COS(RADIANS(latpoint))))
      AND longpoint + (r / (69 * COS(RADIANS(latpoint))))
  ) d
 WHERE distance_in_mi <= r
 ORDER BY distance_in_mi;

邮政编码半径搜索POINT列

2 个答案: