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题解 | 各个视频的平均完播率

select tg.video_id, round( avg(case when timestampdiff(second,start_time,end_time)>=duration then 1 else 0 end),3) as avg_comp_play_rate from tb_u...

2025-03-21

0 34

题解 | 试卷发布当天作答人数和平均分

with tiaojian as ( select uid from user_info where level>5 ),tiaojian1 as ( select exam_id, release_time from examination_info where tag="SQ...

2025-03-18

0 43

题解 | 月均完成试卷数不小于3的用户爱作答的类别

with tiaojian as ( select uid from exam_record group by uid having count(submit_time)/count(distinct month(submit_time))>=3 ) select tag, count(...

2025-03-18

0 52

题解 | 查询连续登陆的用户

with tiaojian as ( select user_id, date(log_time) as pday, row_number()over(partition by user_id order by date(log_time) ) as py from login_tb ),tiao...

2025-03-10

0 33

题解 | 商品销售总额分布

select case when pay_method="" then "error" else pay_method end as py, count(pay_method) as cnt from user_client_log lg left join ...

2025-02-21

0 48

题解 | 商品销售排名

with tiaojian as ( select lg.product_id, type, product_name, sum(price) as py, dense_rank()over(order by sum(price) desc,product_name asc) as pdiff f...

2025-02-21

1 48

题解 | 商品价格排名

#1.先求出每个商品类型的价格的前2名的商品。 #2.在将上述的商品在进行个排序，取前三。 with tiaojian as ( select product_id, product_name, type, price, dense_rank()over(partition by type orde...

2025-02-21

0 73

题解 | 用户购买次数前三

select uid, cnt from( select uid, count(distinct trace_id) as cnt, row_number()over(order by count(distinct trace_id) desc,uid) as pdiff from user_cl...

2025-02-21

1 47

题解 | 下单最多的商品

select t1.product_id, pdiff as cnt from( select product_id, pdiff, dense_rank()over(order by pdiff desc,product_id asc) as py from( select product_...

2025-02-20

0 67

题解 | 完成员工考核试卷突出的非领导员工

with tiaojian as ( select ed.emp_id, eo.exam_id, emp_level, eo.tag, ed.score, timestampdiff(minute,start_time,submit_time) as pt, avg(score)over(part...

2025-02-20

0 48