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MySQL游标循环查找最小值

转载 作者:行者123 更新时间:2023-11-29 03:17:01 25 4
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这是我要完成的任务:

查找指定位置最近的可用车辆。

我有一个 vehicle 表和一个 location 表,如下所示:

CREATE TABLE location
(location_id numeric(8,0) UNIQUE NOT NULL,
address varchar(100),
latitude float,
longitude float,
PRIMARY KEY(location_id)
);

CREATE TABLE vehicle
(license_plate char(6) UNIQUE NOT NULL,
make varchar(30) NOT NULL,
model varchar(30) NOT NULL,
year numeric(4,0) NOT NULL CHECK(year>1990),
state char(2) NOT NULL,
capacity int NOT NULL,
last_location numeric(8,0) DEFAULT NULL,
FOREIGN KEY(last_location) REFERENCES location(location_id) ON DELETE
CASCADE ON UPDATE CASCADE,
PRIMARY KEY(license_plate)
);

我编写了一个查询,它调用一个函数来遍历 vehicle 表,计算与给定位置的距离,并返回具有最小距离的汽车的 license_plate .

SELECT @locationA := 11111111;
SET @loc_lat = (SELECT latitude FROM location WHERE location_id =
@locationA);
SET @loc_long = (SELECT longitude FROM location WHERE location_id =
@locationA);
SELECT license_plate, make, model FROM vehicle
WHERE license_plate = find_car(@loc_lat, @loc_long);

DELIMITER $$
CREATE FUNCTION find_car(loc_lat float, loc_long float) RETURNS char
BEGIN
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
DECLARE CUR_DIST float DEFAULT 1000000000;
DECLARE car_lat NUMERIC;
DECLARE car_long NUMERIC;
DECLARE dist float;
DECLARE closest_car char(6);
DECLARE car_temp char(6);
DECLARE loc_temp numeric;
DECLARE car_cursor CURSOR FOR SELECT license_plate, last_location FROM
vehicle;
OPEN car_cursor;
car_loop: LOOP
FETCH car_cursor INTO car_temp, loc_temp;
SET car_lat = (SELECT latitude FROM location WHERE location_id =
loc_temp);
SET car_long = (SELECT longitude FROM location WHERE location_id =
loc_temp);
SET dist = (SELECT ST_Distance_Sphere(
point(loc_lat, loc_long),
point(car_lat, car_long)
) * .000621371192);
IF dist < CUR_DIST THEN
SET closest_car = car_temp;
SET CUR_DIST = dist;
END IF;
END LOOP;
CLOSE car_cursor;
RETURN(closest_car);
END $$
DELIMITER ;

现在,这没有返回任何内容,我不太清楚为什么。我对 SQL 还是很陌生,所以提前致谢!

最佳答案

我的建议是首先尝试基于集合的方法,看看它的表现如何。

Some good insights on why found here .

一般策略:对于每个location_id,找到下一个最接近的location_id

一些具体策略:

  • 使用jpgunter's GETDISTANCE 函数(代码见下方)
  • Click here在上下文中查看他的代码
  • 使用自连接到 location 表计算所有 location_id 值之间的距离
  • 确定“太远”的任意距离并排除超出该距离的所有内容
  • 这有助于提高性能
  • 选择距离结果最小的

这是一个用于自连接到 location 表和“太远”标准的起始脚本...

SELECT l1.location_id as l1_location_id
,l1.latitude as l1_latitude
,l1.longitude as l1_longitude
,l2.location_id as l2_location_id
,l2.latitude as l2_latitude
,l2.longitude as l2_longitude
,GETDISTANCE(l1.latitude, l1.longitude, l2.latitude, l2.longitude) as l1_12_distance
FROM location AS l1
JOIN location AS l2 ON l1.location_id <> l2.location_id
WHERE GETDISTANCE(l1.latitude, l1.longitude, l2.latitude, l2.longitude) <= 1000; -- JJAUSSI: arbitrary "too far"

这是 jpgunter's GETDISTANCE 函数...

DELIMITER $$
/*
Takes two latitudes and longitudes in degrees. You could comment out the conversion if you want to pass as radians.
Calculate the distance in miles, change the radius to the earth's radius in km to get km.
*/

DROP FUNCTION IF EXISTS GETDISTANCE$$
CREATE FUNCTION GETDISTANCE
(deg_lat1 FLOAT, deg_lng1 FLOAT, deg_lat2 FLOAT, deg_lng2 FLOAT)
RETURNS FLOAT
DETERMINISTIC
BEGIN
DECLARE distance FLOAT;
DECLARE delta_lat FLOAT;
DECLARE delta_lng FLOAT;
DECLARE lat1 FLOAT;
DECLARE lat2 FLOAT;
DECLARE a FLOAT;

SET distance = 0;

/*Convert degrees to radians and get the variables I need.*/
SET delta_lat = radians(deg_lat2 - deg_lat1);
SET delta_lng = radians(deg_lng2 - deg_lng1);
SET lat1 = radians(deg_lat1);
SET lat2 = radians(deg_lat2);

/*Formula found here: http://www.movable-type.co.uk/scripts/latlong.html*/
SET a = sin(delta_lat/2.0) * sin(delta_lat/2.0) + sin(delta_lng/2.0) * sin(delta_lng/2.0) * cos(lat1) * cos(lat2);
SET distance = 3956.6 * 2 * atan2(sqrt(a), sqrt(1-a));

RETURN distance;
END$$
DELIMITER ;

您可能需要考虑修改字段名称 year,因为它是 reserved word .

您可能会发现,如果/当您的数据库增长时,这些名称过于笼统:

  • 位置
  • 地址
  • 纬度
  • 经度

不过,我不知道你的数据库。您的表名和字段名可能恰好符合您的需要。希望这对您有所帮助!

关于MySQL游标循环查找最小值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53798705/

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