题解 | #最长连续登录天数#
最长连续登录天数
https://www.nowcoder.com/practice/cb8bc687046e4d32ad38de62c48ad79b
主要思路：
- 利用ROW_NUMBER 对每个user_id登录时长进行升序排序，得到date_order列；
- 利用DATE_SUB 将fdate与date_order相减，得到“伪日期列-date2，若连续登录，date2应相同；
- COUNT计算连续登陆时长 取 MAX。
SELECT DISTINCT user_id, MAX(consec_day) AS max_consec_days
FROM(
SELECT user_id, date2,
COUNT(date2) AS consec_day
FROM(
SELECT *,
DATE_SUB(fdate, INTERVAL date_order day) AS date2
FROM(
SELECT DISTINCT *,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY fdate) AS date_order
FROM tb_dau
WHERE fdate BETWEEN '2023-01-01' AND '2023-01-31'
) AS t
) AS t2
GROUP BY user_id, date2
) AS t3
GROUP BY user_id;
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