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题解 | #从 Customers 表中检索所有的 ID#

select * from Customers 由于结果只有一列，那么使用*就可以，也可以单独写列的名称 mysql默认是不缺分大小写的，但是在配置文件中是可以修改的。

Mysql

2022-08-12

1 273

题解 | #浙大不同难度题目的正确率#

select t2.difficult_level, round(sum(case when t1.result='right' then 1 else 0 end) / count(t1.device_id),4) as correct_rate from user_...

Mysql

2022-08-06

1 194

题解 | #统计复旦用户8月练题情况#

select t1.device_id,t1.university, count(t2.question_id) as question_cnt, sum(case when result='right' then 1 else 0 end) as right_question_cn...

Mysql

2022-07-15

2 245

题解 | #找出每个学校GPA最低的同学#

select device_id,university,gpa from ( select device_id,university,gpa, dense_rank() over(partition by university order by gpa) as ranking ...

Mysql

2022-07-15

0 236

题解 | #计算用户的平均次日留存率#

select round(count(distinct t1.device_id,t1.date)/count(distinct t.device_id,t.date),4) as avg_ret from question_practice_detail t left join question...

Mysql

2022-07-15

21 429

题解 | #计算用户8月每天的练题数量#

select day(date) as day,count(*) as question_cnt from question_practice_detail where year(date)=2021 and month(date)=8 group by day(date) 没有用其他的函数，只用了...

Mysql

2022-07-14

0 259

题解 | #查看不同年龄段的用户明细#

select device_id,gender, case when age >=25 then "25岁及以上" when age >=20 and age < 25 then "20-24岁" when age &l...

Mysql

2022-07-14

0 211

题解 | #计算25岁以上和以下的用户数量#

select case when age >= 25 then "25岁及以上" else "25岁以下" end as age_cut , count(*) as number from user_profile group by case when ...

Mysql

2022-07-14

2 278

题解 | #统计每个用户的平均刷题数#

select t.university,t2.difficult_level, round(count(t1.question_id)/count(distinct t1.device_id),4) as avg_answer_cnt from user_profile t inner jo...

Mysql

2022-07-14

0 172

题解 | #统计每个学校各难度的用户平均刷题数#

select t.university,t2.difficult_level, round(count(t1.question_id)/count(distinct t1.device_id),4) as avg_answer_cnt from user_profile t inner jo...

Mysql

2022-07-14

0 248