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题解 | 行列互换举一反三

select 学号, sum(if(课程 = '语文',成绩,0)) as 语文成绩, sum(if(课程 = '数学',成绩,0)) as 数学成绩 from 成绩表 group by 学号 order by 学号

2025-10-24

0 4

题解 | 行列互换

select 年, sum(if(月 = 1,值,0)) as m1, sum(if(月 = 2,值,0)) as m2, sum(if(月 = 3,值,0)) as m3, sum(if(月 = 4,值,0)) as m4 from cook group by 年 order by 年

2025-10-24

0 7

题解 | 快递量区间分布

with t as (select 客户id, case when count(distinct 运单号) between 0 and 5 then '0-5' when count(distinct 运单号) between 6 and 10 then '6-10' when count(dist...

2025-10-23

0 7

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 8

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 8

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 12

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 11

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 11

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 16

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 13