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1. 连续登录问题变种

• 题目：
  找出恰好连续登录 3 天的用户（不允许更长的连续区间）。
  表结构：user_logs(user_id, login_date)。

• 参考答案：

  WITH ranked_logs AS (SELECT user_id,login_date,ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_date) AS rnFROM user_logs
  ),
  consecutive_groups AS (SELECT user_id,DATE_SUB(login_date, INTERVAL rn DAY) AS grp,MIN(login_date) AS start_date,MAX(login_date) AS end_date,COUNT(*) AS daysFROM ranked_logsGROUP BY user_id, grp
  )
  SELECT user_id, start_date, end_date
  FROM consecutive_groups
  WHERE days = 3;
  

2. 连续未登录问题

• 题目：
  找出用户最长连续未登录天数（假设表中仅记录登录日期）。
  表结构：user_logs(user_id, login_date)。

• 参考答案：

  WITH next_logs AS (SELECT user_id,login_date,LEAD(login_date) OVER (PARTITION BY user_id ORDER BY login_date) AS next_loginFROM user_logs
  )
  SELECT user_id,MAX(DATEDIFF(next_login, login_date) - 1) AS max_consecutive_missing
  FROM next_logs
  WHERE next_login IS NOT NULL
  GROUP BY user_id;
  

二、窗口函数高级应用

3. 移动平均值计算

• 题目：
  计算用户最近 7 天的平均消费金额（滑动窗口）。
  表结构：orders(user_id, order_date, amount)。

• 参考答案：

  SELECT user_id,order_date,AVG(amount) OVER (PARTITION BY user_id ORDER BY order_date RANGE BETWEEN INTERVAL '6 DAY' PRECEDING AND CURRENT ROW) AS rolling_7day_avg
  FROM orders;
  

4. 增长率计算

• 题目：
  计算每个用户月消费金额的环比增长率。
  表结构：orders(user_id, order_date, amount)。

• 参考答案：

  WITH monthly_sales AS (SELECT user_id,DATE_FORMAT(order_date, '%Y-%m') AS month,SUM(amount) AS total_amountFROM ordersGROUP BY user_id, month
  )
  SELECT user_id,month,total_amount,(total_amount / LAG(total_amount) OVER (PARTITION BY user_id ORDER BY month) - 1) * 100 AS growth_rate
  FROM monthly_sales;
  

三、时间序列分析

5. 缺失日期填充

• 题目：
  生成用户每日登录状态（0 = 未登录，1 = 登录），包括缺失的日期。
  表结构：user_logs(user_id, login_date)。

• 参考答案：

  WITH date_range AS (SELECT user_id,MIN(login_date) AS start_date,MAX(login_date) AS end_dateFROM user_logsGROUP BY user_id
  ),
  all_dates AS (SELECT dr.user_id,d.calendar_dateFROM date_range drCROSS JOIN (SELECT CURDATE() - INTERVAL n DAY AS calendar_dateFROM (SELECT @row := @row + 1 AS n FROM (SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3) t1,(SELECT 0 UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3) t2,(SELECT @row := -1) t3) t) dWHERE d.calendar_date BETWEEN dr.start_date AND dr.end_date
  )
  SELECT ad.user_id,ad.calendar_date,IF(ul.login_date IS NULL, 0, 1) AS is_logged_in
  FROM all_dates ad
  LEFT JOIN user_logs ul 
  ON ad.user_id = ul.user_id AND ad.calendar_date = ul.login_date;
  

6. 周期性检测

• 题目：
  找出用户每周固定某天登录的行为模式（如每周一登录）。
  表结构：user_logs(user_id, login_date)。

• 参考答案：

  WITH day_of_week AS (SELECT user_id,login_date,DAYOFWEEK(login_date) AS dowFROM user_logs
  )
  SELECT user_id,dow,COUNT(DISTINCT WEEK(login_date)) AS weeks_count,COUNT(*) AS login_count
  FROM day_of_week
  GROUP BY user_id, dow
  HAVING login_count = weeks_count; -- 每周该天均登录
  

四、复杂业务场景

7. 购买间隔分析

• 题目：
  计算用户平均购买间隔，并找出间隔超过 30 天的用户。
  表结构：orders(user_id, order_date)。

• 参考答案：

  WITH order_intervals AS (SELECT user_id,order_date,DATEDIFF(order_date, LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date)) AS days_since_lastFROM orders
  )
  SELECT user_id,AVG(days_since_last) AS avg_interval
  FROM order_intervals
  WHERE days_since_last IS NOT NULL
  GROUP BY user_id
  HAVING avg_interval > 30;
  

8. 活跃 / 流失用户分析

• 题目：
  标记用户每月状态（活跃 = 当月有登录，流失 = 连续 3 个月未登录）。
  表结构：user_logs(user_id, login_date)。

• 参考答案：

  WITH months AS (SELECT user_id,DATE_FORMAT(login_date, '%Y-%m') AS month,MAX(login_date) AS last_loginFROM user_logsGROUP BY user_id, month
  ),
  status AS (SELECT m.user_id,m.month,m.last_login,LEAD(m.last_login, 3) OVER (PARTITION BY m.user_id ORDER BY m.month) AS next_3rd_month_loginFROM months m
  )
  SELECT user_id,month,CASE WHEN next_3rd_month_login IS NULL THEN '流失'ELSE '活跃'END AS status
  FROM status;
  

五、进阶挑战

9. 最长连续事件链

• 题目：
  找出用户最长的连续事件链（如连续点赞、评论等，事件类型相同）。
  表结构：events(user_id, event_time, event_type)。

• 参考答案：

  WITH ranked_events AS (SELECT user_id,event_time,event_type,ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY event_time) AS rnFROM events
  ),
  event_groups AS (SELECT user_id,event_type,DATE_SUB(event_time, INTERVAL rn SECOND) AS grp,COUNT(*) AS chain_lengthFROM ranked_eventsGROUP BY user_id, event_type, grp
  )
  SELECT user_id,event_type,MAX(chain_length) AS max_chain
  FROM event_groups
  GROUP BY user_id, event_type;
  

10. 会话识别

• 题目：
  将用户行为按会话分组（假设会话间隔为 30 分钟）。
  表结构：actions(user_id, action_time, action_type)。

• 参考答案：

  WITH time_diff AS (SELECT user_id,action_time,action_type,TIMESTAMPDIFF(MINUTE, LAG(action_time) OVER (PARTITION BY user_id ORDER BY action_time), action_time) AS minutes_since_lastFROM actions
  ),
  session_markers AS (SELECT user_id,action_time,action_type,IF(minutes_since_last > 30 OR minutes_since_last IS NULL, 1, 0) AS new_sessionFROM time_diff
  ),
  sessions AS (SELECT user_id,action_time,action_type,SUM(new_session) OVER (PARTITION BY user_id ORDER BY action_time) AS session_idFROM session_markers
  )
  SELECT * FROM sessions;
  

1. 先手动模拟数据：创建测试表并插入少量数据，验证逻辑正确性。
2. 对比不同方法：例如连续值问题，尝试用 LEAD()、DATE_SUB + ROW_NUMBER 等多种方法实现。
3. 注意边界条件：处理空值、同一天多次记录、跨年 / 跨月等场景。