【SQL每日一练】统计复旦用户8月练题情况

文章目录

题目
一、分析
二、题解
- 1.使用case...when..then
- 2.使用if

题目

现在运营想要了解复旦大学的每个用户在8月份练习的总题目数和回答正确的题目数情况，请取出相应明细数据，对于在8月份没有练习过的用户，答题数结果返回0.

示例代码：

复制代码

drop table if exists `user_profile`;
drop table if  exists `question_practice_detail`;
drop table if  exists `question_detail`;
CREATE TABLE `user_profile` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`gender` varchar(14) NOT NULL,
`age` int ,
`university` varchar(32) NOT NULL,
`gpa` float,
`active_days_within_30` int ,
`question_cnt` int ,
`answer_cnt` int 
);
CREATE TABLE `question_practice_detail` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`question_id`int NOT NULL,
`result` varchar(32) NOT NULL,
`date` date NOT NULL
);
CREATE TABLE `question_detail` (
`id` int NOT NULL,
`question_id`int NOT NULL,
`difficult_level` varchar(32) NOT NULL
);

INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12);
INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25);
INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30);
INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2);
INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70);
INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13);
INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52);
INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong','2021-05-03');
INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong','2021-05-09');
INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong','2021-06-15');
INSERT INTO question_practice_detail VALUES(4,6543,111,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(5,2315,115,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(6,2315,116,'right','2021-08-14');
INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(8,3214,112,'wrong','2021-05-09');
INSERT INTO question_practice_detail VALUES(9,3214,113,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(10,6543,111,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(11,2315,115,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(12,2315,116,'right','2021-08-14');
INSERT INTO question_practice_detail VALUES(13,2315,117,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(14,3214,112,'wrong','2021-08-16');
INSERT INTO question_practice_detail VALUES(15,3214,113,'wrong','2021-08-18');
INSERT INTO question_practice_detail VALUES(16,6543,111,'right','2021-08-13');
INSERT INTO question_detail VALUES(1,111,'hard');
INSERT INTO question_detail VALUES(2,112,'medium');
INSERT INTO question_detail VALUES(3,113,'easy');
INSERT INTO question_detail VALUES(4,115,'easy');
INSERT INTO question_detail VALUES(5,116,'medium');
INSERT INTO question_detail VALUES(6,117,'easy');

提示：以下是本篇文章正文内容，下面案例可供参考

一、分析

1、用where字段操作 WHERE t1.university = '复旦大学'

2、时间方面要注意从题意"8月没有练习过的用户"可知没有答过题的也要统计，

3、但是没有答过题的在左连接下question_id为null 所以要加上

OR b.date IS NULL

4、result字段是字符型的，题目要求统计回答正确的题数，直接计数肯定不行

所以用case 或者 if 函数转换一下，然后用求和函数统计，可以一并把null和

wrong值转换成0值

5、接下来就是统计题目数，和对答题结果求和

6、最后根据用户分一下组

二、题解

1.使用case...when...then

代码如下：

c 复制代码

select
    a.device_id,
    a.university,
    count(b.question_id) as question_cnt,
    sum(case when b.result = 'right' then 1
    else 0
    end) as right_question_cnt
From
    user_profile a
    left join question_practice_detail b on a.device_id = b.device_id
where
    a.university = '复旦大学'
    and (Month (b.date) = 8 or b.date is null)
group by
    a.device_id

2.使用if

代码如下：

c 复制代码

select up.device_id, '复旦大学' as university,
    count(question_id) as question_cnt,
    sum(if(qpd.result='right', 1, 0)) as right_question_cnt
from user_profile as up
 
left join question_practice_detail as qpd
  on qpd.device_id = up.device_id and month(qpd.date) = 8
 
where up.university = '复旦大学'
group by up.device_id