未完成试卷数大于1的有效用户

现有试卷作答记录表exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分），示例数据如下：

id	uid	exam_id	start_time	submit_time	score
1	1001	9001	2021-07-02 09:01:01	2021-07-02 09:21:01	80
2	1002	9001	2021-09-05 19:01:01	2021-09-05 19:40:01	81
3	1002	9002	2021-09-02 12:01:01	(NULL)	(NULL)
4	1002	9003	2021-09-01 12:01:01	(NULL)	(NULL)
5	1002	9001	2021-07-02 19:01:01	2021-07-02 19:30:01	82
6	1002	9002	2021-07-05 18:01:01	2021-07-05 18:59:02	90
7	1003	9002	2021-07-06 12:01:01	(NULL)	(NULL)
8	1003	9003	2021-09-07 10:01:01	2021-09-07 10:31:01	86
9	1004	9003	2021-09-06 12:01:01	(NULL)	(NULL)
10	1002	9003	2021-09-01 12:01:01	2021-09-01 12:31:01	81
11	1005	9001	2021-09-01 12:01:01	2021-09-01 12:31:01	88
12	1005	9002	2021-09-01 12:01:01	2021-09-01 12:31:01	88
13	1006	9002	2021-09-02 12:11:01	2021-09-02 12:31:01	89

还有一张试卷信息表examination_info（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间），示例数据如下：

id	exam_id	tag	difficulty	duration	release_time
1	9001	SQL	hard	60	2020-01-01 10:00:00
2	9002	SQL	easy	60	2020-02-01 10:00:00
3	9003	算法	medium	80	2020-08-02 10:00:00

请统计2021年每个未完成试卷作答数大于1的有效用户的数据（有效用户指完成试卷作答数至少为1且未完成数小于5），输出用户ID、未完成试卷作答数、完成试卷作答数、作答过的试卷tag集合，按未完成试卷数量由多到少排序。示例数据的输出结果如下：

uid	incomplete_cnt	complete_cnt	detail
1002	2	4	2021-09-01:算法;2021-07-02:SQL;2021-09-02:SQL;2021-09-05:SQL;2021-07-05:SQL

解释：2021年的作答记录中，除了1004，其他用户均满足有效用户定义，但只有1002未完成试卷数大于1，因此只输出1002，detail中是1002作答过的试卷{日期:tag}集合，日期和tag间用:连接，多元素间用;连接。

方法一：

select
uid,
sum(incomplete) as incomplete_cnt,
sum(complete) as complete_cnt,
group_concat(distinct predetail separator ';') as detail
from
(
select uid,er.exam_id,start_time,submit_time,
(case when submit_time is null then 1 else 0 end) as incomplete,
(case when submit_time is not null then 1 else 0 end) as complete,
concat(left(start_time,10),':',tag) as predetail
from
exam_record er
join examination_info ei
on er.exam_id = ei.exam_id
where year(start_time) = 2021
group by uid,er.exam_id,start_time,submit_time
order by start_time
) t
group by uid
having incomplete_cnt > 1
and incomplete_cnt < 5
and complete_cnt > 0
order by uid desc

注意：如果使用submit_time进行concat结果如下：

如果只使用start_time进行concat如下：

所以需要加上distinct进行去重，去重同一时间点退出重新登录的情况

方法二:

问题分解：

关联作答记录和试卷信息：left join examination_info on using(exam_id)；（题中exam_record中的exam_id在examination_info均存在，所以用left join和inner join效果一样）
筛选2021年的记录：where year(start_time)=2021
获取各用户的tag,start_time及未完成标记和已完成标记，如果该作答记录交卷了则已完成标记为1，未完成标记为0，否则相反：if(submit_time is null, 1, null) as incomplete
按用户分组：group by uid
统计未完成试卷作答数和已完成试卷作答数：count(incomplete) as incomplete_cnt
统计作答过的tag集合：
- 对于每条作答tag，用:连接日期和tag：concat_ws(':', date(start_time), tag)
- 对于一个人（组内）的多条作答，用;连接去重后的作答记录：group_concat(distinct concat_ws(':', date(start_time), tag) SEPARATOR ';')
筛选未完成试卷作答数大于1的有效用户：having complete_cnt >= 1 and incomplete_cnt BETWEEN 2 and 4
- 完成试卷作答数至少为1：complete_cnt >= 1
- 未完成数小于5：incomplete_cnt < 5
- 未完成试卷作答数大于1：incomplete_cnt > 1

SELECT uid, count(incomplete) as incomplete_cnt,
count(complete) as complete_cnt,
group_concat(distinct concat_ws(':', date(start_time), tag) SEPARATOR ';') as detail
from (
SELECT uid, tag, start_time,
if(submit_time is null, 1, null) as incomplete,
if(submit_time is null, null, 1) as complete
from exam_record
left join examination_info using(exam_id)
where year(start_time)=2021
) as exam_complete_rec
group by uid
having complete_cnt >= 1 and incomplete_cnt BETWEEN 2 and 4
order by incomplete_cnt DESC

方法三：

STEP1:把2021年的数据筛选出来

 
select *
from exam_record er
left join examination_info ei
on er.exam_id=ei.exam_id
where year(er.start_time)=2021

STEP2:按uid聚合，计算incomplete_cnt和complete_cnt

 



select a.uid,
SUM(CASE when a.submit_time is null then 1 END) as incomplete_cnt,
SUM(CASE when a.submit_time is not null then 1 END) as complete_cnt,
GROUP_CONCAT(DISTINCT CONCAT(DATE_FORMAT(a.start_time,'%Y-%m-%d'),':',b.tag) order by start_time SEPARATOR ";") as detail
from exam_record a 
left join examination_info b 
on a.exam_id=b.exam_id
where YEAR(a.start_time)=2021
group by a.uid

STEP3：进行筛选

（1）未完成试卷数大于1 incomplete_cnt>1
（2）完成试卷数至少为1 complete_cnt>=1

（3）未完成数小于5 incomplete_cnt<5

STEP4:将日期和tag组合，函数group_concat([DISTINCT] 要连接的字段 [Order BY ASC/DESC 排序字段] [Separator'分隔符'])

综上：

select er.uid,
count(case when er.submit_time is null then er.start_time else null end) incomplete_cnt,
count(case when er.submit_time is not null then er.start_time else null end) complete_cnt,
GROUP_CONCAT(DISTINCT DATE_FORMAT(er.start_time,'%Y-%m-%d'),':',ei.tag separator ';') detail
from exam_record er
left join examination_info ei
on er.exam_id=ei.exam_id
where year(er.start_time)=2021
group by er.uid
having complete_cnt>=1 and incomplete_cnt<5 and incomplete_cnt>1
order by incomplete_cnt desc;