亲宝软件园·资讯

展开

mysql 判断是否子集 mysql 判断是否为子集的方法步骤

看,月亮在跳舞 人气:3
想了解mysql 判断是否为子集的方法步骤的相关内容吗,看,月亮在跳舞在本文为您仔细讲解mysql 判断是否子集的相关知识和一些Code实例,欢迎阅读和指正,我们先划重点:mysql,判断是否子集,mysql,子集,下面大家一起来学习吧。

一、问题

故事起源于一个查询错漏率的报表:有两个查询结果,分别是报告已经添加的项目和报告应该添加的项目,求报告无遗漏率

何为无遗漏?即,应该添加的项目已经被全部添加

报告无遗漏率也就是无遗漏报告数占报告总数的比率

这里以两个报告示例(分别是已全部添加和有遗漏的报告)

首先,查出第一个结果——报告应该添加的项目

SELECT 
     r.id AS 报告ID,m.project_id 应添加项目
FROM 
  report r 
  INNER JOIN application a ON r.app_id=a.id
  INNER JOIN application_sample s ON a.id=s.app_id
  RIGHT JOIN application_sample_item si ON s.id=si.sample_id       
  RIGHT JOIN set_project_mapping m ON si.set_id=m.set_id
WHERE r.id IN ('44930','44927')
ORDER BY r.id,m.project_id;

然后,再查出第二个结果——报告已经添加的项目

SELECT r.id AS 报告ID,i.project_id AS 已添加项目 
FROM report r 
RIGHT JOIN report_item i ON r.id=i.report_id
WHERE r.id IN ('44930','44927');

以上就是我们要比较的结果集,不难看出报告44927是无遗漏的,而44930虽然项目数量一致,但实际是多添加了项目758,缺少了项目112,是有遗漏的报告

二、解决方案

从问题看,显然是一个判断是否为子集的问题。可以分别遍历已添加的项目和应该添加的项目,如果应该添加的项目在已添加的项目中都能匹配上,即代表应该添加的项目是已添加的项目子集,也就是无遗漏。

通过循环遍历比较确实可以解决这个问题,但是SQL中出现笛卡儿积的交叉连接往往意味着开销巨大,查询速度慢,那么有没有办法避免这一问题呢?

方案一:

借助于函数 FIND_IN_SET和GROUP_CONCAT, 首先认识下两个函数

FIND_IN_SET(str,strlist)

FIND_IN_SET 函数返回了需要查询的字符串在目标字符串的位置

GROUP_CONCAT( [distinct] 要连接的字段 [order by 排序字段 asc/desc  ] [separator '分隔符'] )

GROUP_CONCAT()函数可以将多条记录的同一字段的值,拼接成一条记录返回。默认以英文‘,'分割

但是,GROUP_CONCAT()默认长度为1024

所以,如果需要拼接的长度超过1024将会导致截取不全,需要修改长度

SET GLOBAL group_concat_max_len=102400;
SET SESSION group_concat_max_len=102400;

 从上述两个函数介绍中,我们发现FIND_IN_SET和GROUP_CONCAT都以英文‘,'分割(加粗标识)

所以,我们可以用GROUP_CONCAT将已添加项目的项目连接为一个字符串,然后再用FIND_IN_SET逐一查询应添加项目是否都存在于字符串

1、修改问题中描述中的SQL,用GROUP_CONCAT将已添加项目的项目连接为一个字符串

SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,'') AS 已添加项目列表 
FROM report r 
LEFT JOIN report_item i ON r.id=i.report_id
WHERE r.id IN ('44930','44927')
GROUP BY r.id;

2、用FIND_IN_SET逐一查询应添加项目是否都存在于字符串

SELECT Q.id,FIND_IN_SET(W.应添加项目列表,Q.已添加项目列表) AS 是否遗漏
   FROM 
   (
   -- 报告已经添加的项目 
      SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,'') AS 已添加项目列表 
      FROM report r 
      LEFT JOIN report_item i ON r.id=i.report_id
      WHERE r.id IN ('44930','44927')
      GROUP BY r.id
   )Q,
   (
   -- 报告应该添加的项目 
      SELECT 
         r.id,s.app_id,m.project_id 应添加项目列表
      FROM 
         report r 
         INNER JOIN application a ON r.app_id=a.id
         INNER JOIN application_sample s ON a.id=s.app_id
         INNER JOIN application_sample_item si ON s.id=si.sample_id       
         INNER JOIN set_project_mapping m ON si.set_id=m.set_id
      WHERE r.id IN ('44930','44927')
      ORDER BY r.id,m.project_id
   )W
   WHERE Q.id=W.id;

3、过滤掉有遗漏的报告

 SELECT Q.id,CASE WHEN FIND_IN_SET(W.应添加项目列表,Q.已添加项目列表)>0 THEN 1 ELSE 0 END AS 是否遗漏
   FROM 
   (
   -- 报告已经添加的项目 
      SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,'') AS 已添加项目列表 
      FROM report r 
      LEFT JOIN report_item i ON r.id=i.report_id
      WHERE r.id IN ('44930','44927')
      GROUP BY r.id
   )Q,
   (
   -- 报告应该添加的项目 
      SELECT 
         r.id,s.app_id,m.project_id 应添加项目列表
      FROM 
         report r 
         INNER JOIN application a ON r.app_id=a.id
         INNER JOIN application_sample s ON a.id=s.app_id
         INNER JOIN application_sample_item si ON s.id=si.sample_id       
         INNER JOIN set_project_mapping m ON si.set_id=m.set_id
      WHERE r.id IN ('44930','44927')
      ORDER BY r.id,m.project_id
   )W
   WHERE Q.id=W.id
   GROUP BY Q.id
   HAVING COUNT(`是否遗漏`)=SUM(`是否遗漏`);

4、我们的最终目标是求无遗漏率

 SELECT COUNT(X.id) 无遗漏报告数,Y.total 报告总数, CONCAT(FORMAT(COUNT(X.id)/Y.total*100,2),'%') AS 项目无遗漏率 FROM 
(
  SELECT Q.id,CASE WHEN FIND_IN_SET(W.应添加项目列表,Q.已添加项目列表)>0 THEN 1 ELSE 0 END AS 是否遗漏
   FROM 
   (
   -- 报告已经添加的项目 
      SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,'') AS 已添加项目列表 
      FROM report r 
      LEFT JOIN report_item i ON r.id=i.report_id
      WHERE r.id IN ('44930','44927')
      GROUP BY r.id
   )Q,
   (
   -- 报告应该添加的项目 
      SELECT 
         r.id,s.app_id,m.project_id 应添加项目列表
       FROM 
         report r 
         INNER JOIN application a ON r.app_id=a.id
         INNER JOIN application_sample s ON a.id=s.app_id
         INNER JOIN application_sample_item si ON s.id=si.sample_id       
         INNER JOIN set_project_mapping m ON si.set_id=m.set_id
       WHERE r.id IN ('44930','44927')
    ORDER BY r.id,m.project_id
   )W
   WHERE Q.id=W.id
   GROUP BY Q.id
   HAVING COUNT(`是否遗漏`)=SUM(`是否遗漏`)
 )X,
 (
    -- 总报告数
    SELECT COUNT(E.nums) AS total FROM
    (
      SELECT COUNT(r.id) AS nums FROM report r 
      WHERE r.id IN ('44930','44927')
      GROUP BY r.id
    )E    
 )Y 
 ;

方案二:

上述方案一虽然避免了逐行遍历对比,但本质上还是对项目的逐一对比,那么有没有什么方式可以不用对比呢?

答案当然是有的。我们可以根据统计数量判断是否完全包含。

1、使用union all 将已添加项目与应添加项目联表,不去重

 (
 -- 应该添加的项目
SELECT 
  r.id,m.project_id
FROM 
   report r 
INNER JOIN application a ON r.app_id=a.id
INNER JOIN application_sample s ON a.id=s.app_id
INNER JOIN application_sample_item si ON s.id=si.sample_id       
INNER JOIN set_project_mapping m ON si.set_id=m.set_id
WHERE r.id IN ('44930','44927')
ORDER BY r.id,m.project_id
)
UNION ALL
(
 -- 已经添加的项目
select r.id,i.project_id from report r,report_item i 
where r.id = i.report_id and r.id IN ('44930','44927')
group by r.app_id,i.project_id
 )

从结果可以看出,项目同一个报告下有重复的项目,分别代表了应该添加和已经添加的项目

2、根据联表结果,统计报告重合的项目数量

# 应该添加与已经添加的项目重叠数量
select tt.id,count(*) count from 
(
   select t.id,t.project_id,count(*) from 
   (
      (
        -- 应该添加的项目
        SELECT 
          r.id,m.project_id
        FROM 
          report r 
          INNER JOIN application a ON r.app_id=a.id
          INNER JOIN application_sample s ON a.id=s.app_id
          INNER JOIN application_sample_item si ON s.id=si.sample_id       
          INNER JOIN set_project_mapping m ON si.set_id=m.set_id
        WHERE r.id IN ('44930','44927')
        ORDER BY r.id,m.project_id
      )
      UNION ALL
      (
        -- 已经添加的项目
        select r.id,i.project_id from report r,report_item i 
        where r.id = i.report_id and r.id IN ('44930','44927')
        group by r.app_id,i.project_id
      )
      
   ) t
   GROUP BY t.id,t.project_id
   HAVING count(*) >1 
) tt group by tt.id 

3、将第二步的数量与应该添加的数量作比较,如果相等,则代表无遗漏

select bb.id,aa.count 已添加,bb.count 需添加,
    CASE WHEN aa.count/bb.count=1 THEN 1
    ELSE 0
    END AS '是否遗漏' 
from 
(
# 应该添加与已经添加的项目重叠数量
select tt.id,count(*) count from 
(
   select t.id,t.project_id,count(*) from 
   (
      (
        -- 应该添加的项目
        SELECT 
          r.id,m.project_id
        FROM 
          report r 
          INNER JOIN application a ON r.app_id=a.id
          INNER JOIN application_sample s ON a.id=s.app_id
          INNER JOIN application_sample_item si ON s.id=si.sample_id       
          INNER JOIN set_project_mapping m ON si.set_id=m.set_id
        WHERE r.id IN ('44930','44927')
        ORDER BY r.id,m.project_id
      )
      UNION ALL
      (
        -- 已经添加的项目
        select r.id,i.project_id from report r,report_item i 
        where r.id = i.report_id and r.id IN ('44930','44927')
        group by r.app_id,i.project_id
      )
      
   ) t
   GROUP BY t.id,t.project_id
   HAVING count(*) >1 
) tt group by tt.id 
) aa RIGHT JOIN
(
  -- 应该添加的项目数量
  SELECT 
    r.id,s.app_id,COUNT(m.project_id) count
  FROM 
    report r 
    INNER JOIN application a ON r.app_id=a.id
    INNER JOIN application_sample s ON a.id=s.app_id
    INNER JOIN application_sample_item si ON s.id=si.sample_id       
    INNER JOIN set_project_mapping m ON si.set_id=m.set_id
  WHERE r.id IN ('44930','44927')
  GROUP BY r.id
  ORDER BY r.id,m.project_id
) bb ON aa.id = bb.id 
ORDER BY aa.id

4、求出无遗漏率

select 
    SUM(asr.`是否遗漏`) AS 无遗漏数,COUNT(asr.id) AS 总数,CONCAT(FORMAT(SUM(asr.`是否遗漏`)/COUNT(asr.id)*100,5),'%') AS 报告无遗漏率
from 
(
  select bb.id,aa.count 已添加,bb.count 需添加,
      CASE WHEN aa.count/bb.count=1 THEN 1
      ELSE 0
      END AS '是否遗漏' 
  from 
  (
  # 应该添加与已经添加的项目重叠数量
  select tt.id,count(*) count from 
  (
     select t.id,t.project_id,count(*) from 
     (
        (
          -- 应该添加的项目
          SELECT 
            r.id,m.project_id
          FROM 
            report r 
            INNER JOIN application a ON r.app_id=a.id
            INNER JOIN application_sample s ON a.id=s.app_id
            INNER JOIN application_sample_item si ON s.id=si.sample_id       
            INNER JOIN set_project_mapping m ON si.set_id=m.set_id
          WHERE r.id IN ('44930','44927')
          ORDER BY r.id,m.project_id
        )
        UNION ALL
        (
          -- 已经添加的项目
          select r.id,i.project_id from report r,report_item i 
          where r.id = i.report_id and r.id IN ('44930','44927')
          group by r.app_id,i.project_id
        )
        
     ) t
     GROUP BY t.id,t.project_id
     HAVING count(*) >1 
  ) tt group by tt.id 
  ) aa RIGHT JOIN
  (
    -- 应该添加的项目数量
    SELECT 
      r.id,s.app_id,COUNT(m.project_id) count
    FROM 
      report r 
      INNER JOIN application a ON r.app_id=a.id
      INNER JOIN application_sample s ON a.id=s.app_id
      INNER JOIN application_sample_item si ON s.id=si.sample_id       
      INNER JOIN set_project_mapping m ON si.set_id=m.set_id
    WHERE r.id IN ('44930','44927')
    GROUP BY r.id
    ORDER BY r.id,m.project_id
  ) bb ON aa.id = bb.id 
  ORDER BY aa.id
) asr;

加载全部内容

相关教程
猜你喜欢
用户评论