营销型网站建设菲凡网百度竞价推广登录入口
满足条件的用户的试卷完成数和题目练习数_牛客题霸_牛客网
0 问题描述
基于用户信息表user_info、试卷信息表examination_info、试卷作答记录表exam_record、题目练习记录表practice_record，筛选出 高难度SQL试卷得分平均值大于80并且是7级的用户，统计他们2021年试卷总完成次数和题目总练习次数，结果按试卷完成数升序，按题目练习数降序。
1 数据准备
drop table if exists examination_info,user_info,exam_record,practice_record;
CREATE TABLE examination_info (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',exam_id int UNIQUE NOT NULL COMMENT '试卷ID',tag varchar(32) COMMENT '类别标签',difficulty varchar(8) COMMENT '难度',duration int NOT NULL COMMENT '时长',release_time datetime COMMENT '发布时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;CREATE TABLE user_info (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',uid int UNIQUE NOT NULL COMMENT '用户ID',`nick_name` varchar(64) COMMENT '昵称',achievement int COMMENT '成就值',level int COMMENT '用户等级',job varchar(32) COMMENT '职业方向',register_time datetime COMMENT '注册时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;CREATE TABLE practice_record (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',uid int NOT NULL COMMENT '用户ID',question_id int NOT NULL COMMENT '题目ID',submit_time datetime COMMENT '提交时间',score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;CREATE TABLE exam_record (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',uid int NOT NULL COMMENT '用户ID',exam_id int NOT NULL COMMENT '试卷ID',start_time datetime NOT NULL COMMENT '开始时间',submit_time datetime COMMENT '提交时间',score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;INSERT INTO user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES(1001, '牛客1号', 3100, 7, '算法', '2020-01-01 10:00:00'),(1002, '牛客2号', 2300, 7, '算法', '2020-01-01 10:00:00'),(1003, '牛客3号', 2500, 7, '算法', '2020-01-01 10:00:00'),(1004, '牛客4号', 1200, 5, '算法', '2020-01-01 10:00:00'),(1005, '牛客5号', 1600, 6, 'C++', '2020-01-01 10:00:00'),(1006, '牛客6号', 2000, 6, 'C++', '2020-01-01 10:00:00');INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES(9001, 'SQL', 'hard', 60, '2021-09-01 06:00:00'),(9002, 'C++', 'hard', 60, '2021-09-01 06:00:00'),(9003, '算法', 'medium', 80, '2021-09-01 10:00:00');INSERT INTO practice_record(uid,question_id,submit_time,score) VALUES
(1001, 8001, '2021-08-02 11:41:01', 60),
(1002, 8001, '2021-09-02 19:30:01', 50),
(1002, 8001, '2021-09-02 19:20:01', 70),
(1002, 8002, '2021-09-02 19:38:01', 70),
(1004, 8001, '2021-08-02 19:38:01', 70),
(1004, 8002, '2021-08-02 19:48:01', 90),
(1001, 8002, '2021-08-02 19:38:01', 70),
(1004, 8002, '2021-08-02 19:48:01', 90),
(1004, 8002, '2021-08-02 19:58:01', 94),
(1004, 8003, '2021-08-02 19:38:01', 70),
(1004, 8003, '2021-08-02 19:48:01', 90),
(1004, 8003, '2021-08-01 19:38:01', 80);INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-09-01 09:01:01', '2021-09-01 09:31:00', 81),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 81),
(1003, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 86),
(1003, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:51', 89),
(1004, 9001, '2021-09-01 19:01:01', '2021-09-01 19:30:01', 85),
(1005, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:02', 85),
(1006, 9003, '2021-09-07 10:01:01', '2021-09-07 10:21:01', 84),
(1006, 9001, '2021-09-07 10:01:01', '2021-09-07 10:21:01', 80);2 数据分析
select t1.uid,count(distinct case when year(t2.submit_time) = '2021' then t2.id else null end) as exam_cnt, count(distinct case when year(t3.submit_time) = '2021' then t3.id else null end) as question_cnt
from (select uidfrom exam_record where uid in (select uid from user_info where level = 7 ) and exam_id in (select exam_id from examination_info where tag = 'SQL' and difficulty = 'hard')group by uid having sum(score) / count(score) > 80 ) t1
left join exam_record t2 on t1.uid = t2.uid
left join practice_record t3 on t1.uid = t3.uid group by t1.uidorder by exam_cnt asc , question_cnt desc ;
-- 结果按试卷完成数升序，按题目练习数降序
思路分析：
- step1: 先筛选出 平均值大于80并且是7级用户的uid，得到t1
- step2：t1分别与t2、t3关联，要用left join，因为有些uid可能没做某个试卷或练习，也要保留记录
- step3：count(distinct )时，以id区分(case when ..then id )，不能以exam_id区分，因为存在一个uid可能对同一个试卷或练习做过多次。