mysql 判断是否为子集的方法步骤
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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