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
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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;
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然后,再查出第二个结果——报告已经添加的项目
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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' );
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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
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set global group_concat_max_len=102400;
set session group_concat_max_len=102400;
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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
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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;
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2、用find_in_set逐一查询应添加项目是否都存在于字符串
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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;
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3、过滤掉有遗漏的报告
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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`);
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4、我们的最终目标是求无遗漏率
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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
;
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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
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(
-- 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
)
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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
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# 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
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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
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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
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4、求出无遗漏率
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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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原文链接:https://blog.csdn.net/kk_gods/article/details/112894187