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【牛客】SQL128 未完成试卷数大于1的有效用户

2026-11-01 19:55:14 来源:互联网转载

描述

���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:

()
iduidexam_idstart_timesubmit_timescore
1100190012021-07-02 09:01:012021-07-02 09:21:0180
2100290012021-09-05 19:01:012021-09-05 19:40:0181
3100290022021-09-02 12:01:01(NULL)(NULL)
4100290032021-09-01 12:01:01(NULL)(NULL)
5100290012021-07-02 19:01:012021-07-02 19:30:0182
6100290022021-07-05 18:01:012021-07-05 18:59:0290
7100390022021-07-06 12:01:01(NULL)(NULL)
8100390032021-09-07 10:01:012021-09-07 10:31:0186
9100490032021-09-06 12:01:01(NULL)(NULL)
10100290032021-09-01 12:01:012021-09-01 12:31:0181
11100590012021-09-01 12:01:012021-09-01 12:31:0188
12100590022021-09-01 12:01:012021-09-01 12:31:0188
13100690022021-09-02 12:11:012021-09-02 12:31:0189

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:

idexam_idtagdifficultydurationrelease_time
19001SQLhard602020-01-01 10:00:00
29002SQLeasy602020-02-01 10:00:00
39003算法medium802020-08-02 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:

()
uidincomplete_cntcomplete_cntdetail
1002242021-09-01: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。

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 > 0
    • 未完成数小于5:incomplete_cnt
    • 未完成试卷作答数大于1:incomplete_cnt > 1
    with cte as(    select    uid,exam_id,tag,start_time,    if(submit_time is null,0,1) as sub_num    from    exam_record left join examination_info    using(exam_id)    where left(start_time,4)=2021)selectuid,(count(sub_num)-sum(sub_num)) as incomplete_cnt,sum(sub_num) as complete_cnt,group_concat(distinct concat_ws(':', date(start_time), tag) separator ';') as detailfrom ctegroup by uidhaving incomplete_cnt between 2 and 4 and complete_cnt>0order by incomplete_cnt desc

     

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