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题解 | 统计各岗位员工平均工作时长

SELECT S.post, ROUND(AVG(TIMESTAMPDIFF(minute, A.first_clockin, A.last_clockin)/60), 3) AS work_hours FROM staff_tb S INNER JOIN attendent_tb A ...

2025-12-14

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题解 | 推荐内容准确的用户平均评分

select avg(DISTINCT score) avg_score from user_action_tb a join recommend_tb b on a.user_id=b.rec_user where a.hobby_l=b.rec_info_l

2025-12-12

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题解 | 统计各等级会员用户下订单总额

select u.vip, IFNULL(sum(o.order_price), 0) as order_total from uservip_tb u left join order_tb o on u.user_id = o.user_id group by u.vip order...

2025-12-11

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题解 | 查询培训指定课程的员工信息

select s.staff_id,staff_name from staff_tb s join cultivate_tb c on c.staff_id=s.staff_id where course like "%3%" --course like "%3%&qu...

2025-12-10

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题解 | 统计所有课程参加培训人次

select sum(t1) staff_nums from (select case when course is NULL then 0 else length(course)-length(replace(course,",",""))+1 end t...

2025-12-10

0 26

题解 | 查询连续入住多晚的客户信息？

SELECT t.user_id, t.room_id, g.room_type, datediff(checkout_time,checkin_time) days from checkin_tb t join guestroom_tb g on t.room_id...

2025-12-10

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题解 | 获得积分最多的人(三)

select u.id, u.name, a.grade_num from ( select t2.user_id, t2.grade_num, rank() over (order by t2.grade_num desc) as t1 f...

2025-12-06

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题解 | 获得积分最多的人(二)

SELECT U.id AS user_id, U.name AS user_name, T2.grade_num AS grade_num FROM ( SELECT user_id, SUM(grade_num) AS grade_num, DEN...

2025-12-05

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题解 | 获得积分最多的人(一)

SELECT u.name, SUM(g.grade_num) as grade_sum FROM grade_info g left JOIN user u ON g.user_id = u.id GROUP BY u.id, u.name ORDER BY grade_s...

2025-12-05

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题解 | 最差是第几名(二)

select grade FROM (SELECT grade,SUM(number) OVER(ORDER BY grade) T1, SUM(number) OVER(ORDER BY grade DESC) T2, (SELECT SUM(number) FROM class_grade) T...

2025-12-05

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