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题解 | 统计借阅量

with t as (select book_id,book_title,branch_name, date_format(borrow_date,'%Y-%m') as month, year(borrow_date) as year from Books bk left join BorrowR...

2025-10-22

0 5

题解 | 统计借阅量

with t as (select bk.book_id as book_id,book_title,record_id,borrow_date,bc.branch_id as branch_id,branch_name from Books bk left join BorrowRecords b...

2025-10-22

0 6

题解 | 购买行为分析3

select 顾客ID from (select 顾客ID,group_concat(distinct 产品) as 产品集合 from 销售订单表 group by 顾客ID) t where 产品集合 like '%ProductA%' and 产品集合 like '%ProductB%' a...

2025-10-17

0 10

题解 | 查找重复数据

select 姓名 from 学生表 group by 姓名 having count(*) > 1

2025-10-15

0 9

题解 | 购买行为分析

select count(distinct 顾客ID) as 购买人数, round(sum(销售数量*零售价)) as 销售金额, sum(零售价*销售数量)/count(distinct 顾客ID) as 客单价, round(sum(销售数量)/count(distinct 顾客ID)) as...

2025-10-15

0 8

题解 | 连续两次作答试卷的最大时间窗

with t as (select uid,start_time, lag(start_time,1)over(partition by uid order by start_time desc) as next_time from exam_record where year(start_time...

2025-09-26

0 15

题解 | 第二快/慢用时之差大于试卷时长一半的试卷

with t as (select ei.exam_id as exam_id,duration, timestampdiff(second,start_time,submit_time)/60 as time_record, rank()over(partition by exam_id orde...

2025-09-25

0 12

题解 | 第二快/慢用时之差大于试卷时长一半的试卷

with t as (select ei.exam_id as exam_id,duration, timestampdiff(second,start_time,submit_time)/60 as time_record, rank()over(partition by exam_id orde...

2025-09-25

0 18

题解 | 第二快/慢用时之差大于试卷时长一半的试卷

with t as (select ei.exam_id as exam_id,duration, timestampdiff(second,start_time,submit_time)/60 as time_record, rank()over(partition by exam_id orde...

2025-09-25

0 14

题解 | 每类试卷得分前3名

select tag,uid,ranking from ( select tag,uid, rank()over(partition by tag order by max(score) desc,min(score) desc,uid desc) as ranking from examinati...

2025-09-25

0 12