SQL面试题练习 —— 统计最大连续登录天数区间

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 rn
           FROM user_login_tb
           group 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,
             CASE
                 WHEN 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 part
      FROM (SELECT *,
                   DATE_SUB(login_date, rn) AS dd
            FROM t) t1
      GROUP 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