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题解 | 查询连续登陆的用户三种方法

select t3.user_id from ( select user_id, count(*) as cut from ( select user_id, date_s...

2026-04-01

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题解 | 最长连续登录天数三种方法

select c.user_id, max(c.cont_) as max_consec_days from ( select b.user_id, count(b.grp) cont_ from ( select a.user_id, fdate, ...

2026-04-01

0 25

题解 | 最长连续登录天数可以使用recursive

with recursive cte as( select fdate, user_id, 1 as depth from tb_dau union all select t....

2026-04-01

1 22

题解 | 宠物猫繁育族谱追溯与遗传病风险评估使用recursive 就可以解决

with recursive cte as( select b.child_cat_id as descendant_id, c1.cat_name as descendant_name, 1 as depth, b.healt...

2026-03-31

0 31

题解 | SaaS产品高价值用户活跃度分析

select concat(u.user_name, '(', u.user_id, ')') as user_profile, sum( case when ue.event_type = 'create_task' then 5 ...

2026-03-30

0 25

题解 | 游戏平台新玩家消费与进阶行为分析

select p.player_id, p.username, p.level as current_level, case when p.level >= 30 then "高阶玩家" else "新秀玩家" end a...

2026-03-30

0 23

题解 | SaaS平台企业客户新功能采纳度分析

select t.team_id, t.team_name, count(*) as april_usage_count, case when count(*) >50 then "深度采纳团队" else "普通采纳团队&quo...

2026-03-30

0 23

题解 | 在线教育平台活跃学员课程评价分析

select c.user_id, round(avg(cr.rating),2) as average_rating, case when avg(cr.rating)>=4 then '优质反馈学员' else '普通反馈学员' end as feedback_ty...

2026-03-29

1 26

题解 | 智能家居设备高能耗异常监控分析

select d.device_name, replace(upper(d.location), ' ', '_') as location_code, d.total_usage, case when d.total_usage >= 50 t...

2026-03-29

0 21

题解 | 近7天骑手履约时效看板

with t1 as ( select max(delivered_ts) as max_date from parcel ), t2 as( select c.courier_id, c.courier_name, c.cit...

2026-03-29

0 27