SQL语句查询连续N天登录用户(解决方案)

DROP TABLE IF EXISTS t_login_records;#若表存在删除
-- 创建用户登录记录表
CREATE TABLE t_login_records (
    id INT PRIMARY KEY AUTO_INCREMENT,
    user_id INT NOT NULL,
    login_date DATE NOT NULL
)ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci COMMENT='用户登录记录表';

-- 查看表结构	
DESC t_login_records;

INSERT INTO t_login_records (user_id, login_date) VALUES
(1, '2023-01-01'),
(1, '2023-01-02'),
(1, '2023-01-03'),  -- 用户1连续3天登录
(1, '2023-01-05'),
(2, '2023-01-01'),
(2, '2023-01-02'),
(2, '2023-01-04'),  -- 用户2不连续
(3, '2023-01-01'),
(3, '2023-01-02'),
(3, '2023-01-03'),  -- 用户3连续3天登录
(3, '2023-01-04'),  -- 用户3连续4天登录
(4, '2023-01-01'),
(4, '2023-01-03'),
(4, '2023-01-05'),  -- 用户4不连续    
(5, '2023-01-01'),
(5, '2023-01-02'),
(5, '2023-01-03'),  -- 用户5连续3天登录
(5, '2023-01-04'),  -- 用户5连续4天登录
(5, '2023-01-05'),	-- 用户5连续5天登录
(5, '2023-02-01'),	-- 断开
(5, '2023-02-02'),  -- 用户5再连续2天登录
(5, '2023-02-03');  -- 用户5再连续3天登录
SELECT * FROM t_login_records;

-- 方法1：自连接查询
SELECT DISTINCT t1.user_id
FROM t_login_records t1
JOIN t_login_records t2 
  ON t1.user_id = t2.user_id 
  AND DATEDIFF(t2.login_date, t1.login_date) = 1
JOIN t_login_records t3 
  ON t1.user_id = t3.user_id 
  AND DATEDIFF(t3.login_date, t1.login_date) = 2;

-- 方法2：窗口函数（适用于支持LEAD函数的数据库，如MySQL 8.0+、PostgreSQL）
SELECT DISTINCT user_id
FROM (
  SELECT 
    user_id,
    login_date,
    LEAD(login_date, 1) OVER (PARTITION BY user_id ORDER BY login_date) AS next_day,
    LEAD(login_date, 2) OVER (PARTITION BY user_id ORDER BY login_date) AS next_2_days
  FROM t_login_records
) t
WHERE DATEDIFF(next_day, login_date) = 1 
  AND DATEDIFF(next_2_days, login_date) = 2;

-- 方法3：日期差值分组（适用于支持ROW_NUMBER的数据库）
SELECT user_id
FROM (
  SELECT 
    user_id,
    login_date,
    DATE_SUB(login_date, INTERVAL ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_date) DAY) AS grp
  FROM t_login_records
) t
GROUP BY user_id, grp
HAVING COUNT(DISTINCT login_date) >= 3;