SQL语句实现用户连续登录最长天数

-- 方法1：嵌套查询
-- 取rn=1每个用户连续登录最长的天数
-- 同一用户可能有多个连续登录周期，若存在最长周期相同，取距今最近的起始日期和终止日期
SELECT
	user_id,
	consecutive_days AS '连续登录最长天数',
	start_date AS '起始日期',
	end_date AS '终止日期' 
FROM
	(
    -- 用户连续登录天数降序排列，加序号
	SELECT
		*,
		ROW_NUMBER ( ) OVER ( PARTITION BY user_id ORDER BY consecutive_days DESC, nd_date DESC) AS rn
	FROM
		(
        -- 用户连续登录天数：可能有多个，只要断开就重新计算，最小为1
        -- 起始日期、终止日期
		SELECT
			user_id,
			COUNT( DISTINCT login_date ) AS consecutive_days,
			MIN( login_date ) AS start_date,
			MAX( login_date ) AS end_date 
		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 ser_id,grp 
		) a 
	) b 
WHERE
	rn = 1;

WITH ConsecutiveCTE AS (
    SELECT 
        user_id,
        COUNT(DISTINCT login_date) AS consecutive_days,
        MIN(login_date) AS start_date,
        MAX(login_date) AS end_date
    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
)
SELECT 
    user_id,
    consecutive_days AS '连续登录最长天数',
    start_date AS '起始日期',
    end_date AS '终止日期'
FROM (
    SELECT 
        *,
        -- 对每个用户，标记其最大连续天数的记录
        ROW_NUMBER ( ) OVER ( PARTITION BY user_id ORDER BY consecutive_days DESC, nd_date DESC) AS rn
    FROM ConsecutiveCTE
) ranked
WHERE rn = 1; -- 只保留每个用户最大连续天数的记录

WITH [recursive] cte_name [(column_alias1, column_alias2,...)] AS (
    -- CTE的查询语句，通常是一个SELECT查询
    SELECT_statement
)
-- 使用CTE的主查询语句
SELECT * FROM cte_name;