描述
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:用户信息表user_profile
id | device_id | gender | age | university | gpa | active_days_within_30 |
1 | 2138 | male | 21 | 北京大学 | 3.4 | 7 |
2 | 3214 | male | 复旦大学 | 4.0 | 15 | |
3 | 6543 | female | 20 | 北京大学 | 3.2 | 12 |
4 | 2315 | female | 23 | 浙江大学 | 3.6 | 5 |
5 | 5432 | male | 25 | 山东大学 | 3.8 | 20 |
6 | 2131 | male | 28 | 山东大学 | 3.3 | 15 |
7 | 4321 | female | 26 | 复旦大学 | 3.6 | 9 |
示例:question_practice_detail
id | device_id | question_id | result | date |
1 | 2138 | 111 | wrong | 2021-05-03 |
2 | 3214 | 112 | wrong | 2021-05-09 |
3 | 3214 | 113 | wrong | 2021-06-15 |
4 | 6543 | 111 | right | 2021-08-13 |
5 | 2315 | 115 | right | 2021-08-13 |
6 | 2315 | 116 | right | 2021-08-14 |
7 | 2315 | 117 | wrong | 2021-08-15 |
…… |
根据示例,你的查询应返回以下结果:
device_id | university | question_cnt | right_question_cnt |
3214 | 复旦大学 | 3 | 0 |
4321 | 复旦大学 | 0 | 0 |
解:
select up.device_id, up.university,
case when count(question_id) > 0 then count(question_id) else 0 end as question_cnt,
sum(if(result = 'right', 1, 0)) as right_question_cnt
from user_profile up left join question_practice_detail qpd on up.device_id = qpd.device_id and month(date) = 08
where university = '复旦大学'
group by device_id;
函数积累
if(判断式,为真返回值,为假返回值)