tingdasheng 2020-05-24 22:07 采纳率: 0%
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关于mysql存储过程执行时间长,求优化思路?

现有一张包含经(lon)纬(lat)度的A (c_pc_stake_info)表,在另一张B表( b_pc_compaction_info)中也包含有经(lon)纬(lat)度字段,需要返回的数据是A表的经纬度、通过A表每两条数据的经纬度和B表的经纬度相比较,得到B表数据在A表这两条数据经纬度之间的个数

CREATE DEFINER=root@% PROCEDURE compaction(
out num INTEGER,
out startLon DOUBLE,
out startLat DOUBLE,
out startStakeNum varchar(100),
out endLon DOUBLE,
out endLat DOUBLE,
out endStakeNum varchar(100)
)
BEGIN

-- 定义变量
DECLARE s int DEFAULT 0;
DECLARE stakeId INTEGER;
DECLARE lons DOUBLE;
DECLARE lats DOUBLE;
DECLARE stakeNum VARCHAR(50);
DECLARE compaction_lon DOUBLE;
DECLARE compaction_lat DOUBLE;

-- 临时表
-- 封装最后所得数据
create temporary table if not exists stake_info_vo (
number INTEGER,
startLon DOUBLE,
startLat DOUBLE,
startStakeNum VARCHAR(50),
endLon DOUBLE,
endLat DOUBLE,
endStakeNum VARCHAR(50)
);
-- 经B表筛选过的数据插入此表
create temporary table if not exists lon_lat (

lon1 DOUBLE,
lat1 DOUBLE
);

BEGIN
-- B表筛选后的数据
declare lonLat CURSOR for
SELECT
lon,
lat
from b_pc_compaction_info
where 1=1
and SUBSTRING_INDEX(create_time," ",1) = SUBSTRING_INDEX((select max(create_time) from b_pc_compaction_info)," ",1)
-- ;
and LENGTH(SUBSTRING_INDEX(lon,".",-1)) = 8
and LENGTH(SUBSTRING_INDEX(lat,".",-1)) = 8;
-- insert into lon_lat VALUES (compaction_lon,compaction_lat);
DECLARE CONTINUE HANDLER FOR NOT FOUND SET s=1;
open lonLat;
fetch lonLat into compaction_lon,compaction_lat;
set s = 0;
while(s=0) do
insert into lon_lat VALUES (compaction_lon,compaction_lat);
fetch lonLat into compaction_lon,compaction_lat;
end while;
-- 关闭游标
close lonLat;
END;

 BEGIN
-- 定义游标,并将sql结果集赋值到游标中
DECLARE report CURSOR FOR
      -- 查询A表数据

select
id,lon,lat,stake_num
from c_pc_stake_info
ORDER BY id;

-- 声明当游标遍历完后将标志变量置成某个值
DECLARE CONTINUE HANDLER FOR NOT FOUND SET s=1;

-- 打开游标
open report;


    -- 将游标中的值赋值给变量,注意:变量名不要和返回的列名同名,变量顺序要和sql结果列的顺序一致
    fetch report into stakeId,lons,lats,stakeNum;
            set s=0;

    -- 当s不等于1,也就是未遍历完时,会一直循环
    while (s<>1) do

        -- 通过遍历获取A表的第二条数据
      SELECT 
       lon,
       lat,
      stake_num
       into endLon,endLat,endStakeNum
    FROM c_pc_stake_info
    WHERE id = (SELECT MIN(id) FROM c_pc_stake_info WHERE id > stakeId);

            -- 获取B表数据在A表每两条数据经纬度之间的个数
                     SELECT count(1) into num
                     from lon_lat 
                     where 1=1
                 and lon1 between lons and endLon
                     and lat1 between lats and endLat
                ;


            SET startLon  = lons;
     set startLat  = lats;      
     set startStakeNum  = stakeNum;
             -- set num = 1;
     INSERT INTO stake_info_vo VALUES (num,startLon,startLat,startStakeNum,endlon,endlat,endStakeNum);  

        -- 当s等于1时表明遍历以完成,退出循环
        fetch report into stakeId,lons,lats,stakeNum;
    end while;
-- 关闭游标
close report;
    SELECT * from stake_info_vo;
    -- SELECT * from lon_lat;

end;
truncate TABLE stake_info_vo;
truncate TABLE lon_lat;

END

第一次用csdn,问题有些格式不是太对,望大佬们谅解

  • 写回答

1条回答 默认 最新

  • 周祭酒 2020-05-25 08:42
    关注

    如果不改设计的话,应该就是排序算法的选择问题了。
    如果改设计的话,需要的是获取区间的个数,那就尽量让他查到即得到结果。那可以每条数据变化时更新他所属区间这条数据所包含的个数;或者是延时更新,即每5分钟后台统计一次,每次查询获取的是上一个5分钟统计的结果

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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