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全部文章（共26篇）

题解 | 最差是第几名(二)

-- 6: 3 and 4 -- 7: 4 and 4 -- 12: 6 and 7 with temp0 as ( select sum(number) as total_number from class_grade ), temp1 as ( select floor...

2026-02-21

0 32

题解 | 牛客的课程订单分析(五)

with temp0 as ( select user_id,count(*) as cnt from order_info oi where date >= "2025-10-15" and product_name in ("C++&q...

2026-02-20

0 29

题解 | SaaS产品高价值用户活跃度分析

with temp0 as ( select distinct u.user_id from users u inner join user_events ue on u.user_id = ue.user_id where plan_type = "Pro...

2026-02-07

0 40

题解 | 智能家居设备高能耗异常监控分析

with temp0 as ( select category,round(sum(usage_kwh)/count(*),2) as avg_usage_kwh from smart_devices sd inner join energy_logs el on sd.d...

2026-02-07

0 38

题解 | 游戏平台新玩家消费与进阶行为分析

with temp0 as ( select distinct p.player_id from players p inner join transactions t on p.player_id = t.player_id where date_format(cr...

2026-02-03

0 35

题解 | SaaS平台企业客户新功能采纳度分析

with temp0 as ( select team_id,min(date_format(usage_timestamp,"%Y-%m-%d")) as first_ever_usage_date from feature_usage group by...

2026-02-03

0 39

题解 | 最长连续登录天数

with temp0 as ( select user_id,fdate,row_number()over(partition by user_id order by fdate asc) as rk from tb_dau td where fdate >= &quo...

2026-01-11

0 50

题解 | 近7天骑手履约时效看板

with temp0 as ( select max(delivered_ts) as benchmark_day from parcel ), temp1 as ( select p.courier_id, count(*) as orde...

2026-01-10

0 40

题解 | 统计每个产品的销售情况

with temp0 as ( select product_id, date_format(order_date, "%Y-%m") as date, sum(quantity) a...

2025-12-27

0 64

题解 | #查询单日多次下订单的用户信息？#

select order_date, t.user_id, order_nums, vip from ( select user_id, date_format(order_time, "%Y-%...

Mysql

2025-12-23

0 51