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sql练习-面试前刷题

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题解 | 查询连续登陆的用户

select user_id from ( select user_id, count(date_sub_rn) as continue_days from ( select user_id, ...

2025-05-30

0 48

题解 | 统计各岗位员工平均工作时长

select post, round(avg(work_time),3) as work_hours from ( select s_tb.staff_id, s_tb.post, timestampdiff(m...

2025-05-30

0 48

题解 | 每个商品的销售总额

select name as product_name, sum_quantity as total_sales, row_number() over(partition by category order by sum_quantity desc) ...

2025-05-28

0 30

题解 | 推荐内容准确的用户平均评分

select round(sum(score) / count(rec_user),3) avg_score from ( select distinct rc_tb.rec_user, u_a_tb.score from ...

2025-05-28

0 44

题解 | 查询培训指定课程的员工信息

select a.staff_id, st.staff_name from ( select staff_id from cultivate_tb where course lik...

2025-05-28

0 64

题解 | 统计所有课程参加培训人次

select sum(cnt) staff_nums from ( select *, case when length(course) = 7 then 1 when length(course...

2025-05-28

0 50

题解 | 查询连续入住多晚的客户信息？

select * from ( select a.user_id, a.room_id, gr_tb.room_type, datediff(checkout_time,checkin_time) days ...

2025-05-28

0 60

题解 | 获取指定客户每月的消费额

select concat('2023-',every_month) time, sum(t_amount) total from ( select substr(t.t_time,6,2) every_month, t.t_a...

2025-05-27

0 54

题解 | 分析客户逾期情况

select pay_ability, concat(round(count(overdue_days) / count(*) * 100,1),'%') overdue_ratio from loan_tb as lt left join ...

2025-05-27

0 51

题解 | 最长连续登录天数

select user_id, max(cnt) max_consec_days from ( select user_id, sub_fdate, count(*) cnt from ( ...

2025-05-27

0 47