1 题目

2 建表语句

CREATE TABLE IF NOT EXISTS user_login_tb (uid INT,login_date DATE
);
insert into user_login_tb(uid, login_date) values( 1, '2022-08-02'),(1, '2022-08-03'),(2, '2022-08-03'),(2, '2022-08-04'),(2, '2022-08-05'),(2, '2022-08-05'),(3, '2022-08-05'),(2, '2022-08-06'),(4, '2022-08-06'),(4, '2022-08-07');

3 题解

WITH t AS (SELECT uid,login_date,ROW_NUMBER() OVER (PARTITION BY uid ORDER BY login_date) AS rnFROM user_login_tbgroup by uid, login_date -- 一个用户一天可能多次登录
)
SELECT part                AS days_range,COUNT(DISTINCT uid) AS user_num
FROM (SELECT uid,dd,COUNT(*) AS cnd,CASEWHEN COUNT(*) < 2 THEN '未连续登录'WHEN COUNT(*) < 4 THEN '连续登录2~3天'WHEN COUNT(*) < 8 THEN '连续登录4~7天'ELSE '连续登录大于7天'END  AS partFROM (SELECT *,DATE_SUB(login_date, rn) AS ddFROM t) t1GROUP BY uid, dd) t2
GROUP BY part
-- 记住sql的执行顺序
-- FROM -> JOIN -> ON -> WHERE -> GROUP BY -> HAVING -> SELECT -> DISTINCT -> ORDER BY -> LIMIT
ORDER BY user_num DESC;

执行结果