题解 | #大小写混乱时的筛选统计#

大小写混乱时的筛选统计

https://www.nowcoder.com/practice/81cb12931a604811ae69d332515c7734

跑出来的代码。分享思路。

题目要求:

(1)筛选出试卷作答数小于3的类别tag;

(2)输出exam_record中的tag,upper(tag)对应的answer_cnt。也可以理解为把(exam_record表中各类tag和answer_cnt)中的answer_cnt替换成upper(tag)对应的answer_cnt,并保持tag不变。

(2)可能比较难理解

思路是:

创建t,储存exam_record表中各类tag和answer_cnt,为什么examination_info left join exam_record,因为examination_info 表里的tag种类是完整的,也可能后面判断比如(tag的answer_cnt<>0,upper(tag)的answer_cnt不存在)

创建t1,储存①满足条件(answer_cnt<3 且 不全为大写字母)的tag ②把tag转大写,记为u_tag;

创建t2,从t中筛选出u_tag和answer_cnt;

最后连表,用t1 left join t2,因为t1中有tag(题目要求tag),条件answer_cnt<>0,剔除exam_record表没有的tag

with t as (
select any_value(t.exam_id) as exam_id,tag,any_value(count(uid)) as answer_cnt
from examination_info t left join exam_record t1 on t.exam_id=t1.exam_id
group by tag)
,t1 as(
select upper(tag) as u_tag,tag
from t
where answer_cnt<3 and tag<>upper(tag))
,t2 as (
select tag,answer_cnt
from t
where tag in(
    select u_tag from t1
))

select t1.tag,answer_cnt
from t1 left join t2 on t1.u_tag=t2.tag
where answer_cnt <> 0

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