描述
题目:现在运营想要计算出2021年8月每天用户练习题目的数量,请取出相应数据。 示例:question_practice_detailid | 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 |
…… |
day | question_cnt |
13 | 5 |
14 | 2 |
15 | 3 |
16 | 1 |
18 | 1 |
select extract(day from date)day,count(*)question_cnt from question_practice_detail where date in( select date from question_practice_detail where year(date) = 2021 and month(date) = 8 ) group by day
select case when year(date) = 2021 and month(date) = 8 then extract(day from date) else null end day, count(*) question_cnt from question_practice_detail group by day
select extract(day from date) day,count(*)question_cnt from question_practice_detail where date_format(date,"%Y-%m")="2021-08" group by day
select day(date) `day`,count(question_id) question_cnt from question_practice_detail where date_format(date, "%Y-%m")="2021-08" group by day
select day(date) as `day`, count(question_id) question_cnt from question_practice_detail where month(date) = 8 and year(date) = 2021 group by date
- 限定条件:2021年8月,写法有很多种,比如用year/month函数的
year(date)=2021 and month(date)=8
, - 比如用date_format函数的
date_format(date, "%Y-%m")="202108"
- 每天:按天分组
group by date
- 题目数量:count(question_id)