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:
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:
selecta.device_id,a.university,count(b.question_id) as question_cnt,sum(case when b.result = 'right' then 1else 0end) as right_question_cnt
Fromuser_profile aleft join question_practice_detail b on a.device_id = b.device_id
wherea.university = '复旦大学'and (Month (b.date) = 8 or b.date is null)
group bya.device_id
2.使用if
代码如下:
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 upleft join question_practice_detail as qpdon qpd.device_id = up.device_id and month(qpd.date) = 8where up.university = '复旦大学'
group by up.device_id