目录
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)
- str: 需要查询的字符串
- strlist: 参数以英文”,”分隔,如 (1,2,6,8,10,22)
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;
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