Sql 获取最长用户条纹的正确计数

我很难获得最长用户连胜的正确计数。连续天数是指每个用户都有签入的连续天数

任何帮助都将不胜感激。下面是我的脚本和示例数据：

签入表：

user_id  goal_id   check_in_date
------------------------------------------      
| colt | 40365fa0 | 2019-01-07 15:35:53
| colt | d31efe70 | 2019-01-11 15:35:52
| berry| be2fcd50 | 2019-01-12 15:35:51
| colt | e754d050 | 2019-01-13 15:17:16
| colt | 9c87a7f0 | 2019-01-14 15:35:54
| colt | ucgtdes0 | 2019-01-15 12:30:59

    WITH dates(DATE) AS
      (SELECT DISTINCT Cast(check_in_date AS DATE),
                       user_id
       FROM check_ins),
         GROUPS AS
      (SELECT Row_number() OVER (
                                ORDER BY DATE) AS rn, DATE - (Row_number() OVER (ORDER BY DATE) * interval '1' DAY) AS grp, DATE, user_id
       FROM dates)
    SELECT Count(*) AS streak,
           user_id
    FROM GROUPS
    GROUP BY grp,
             user_id
    ORDER BY 1 DESC; 

PostgreSQL脚本：

user_id  goal_id   check_in_date
------------------------------------------      
| colt | 40365fa0 | 2019-01-07 15:35:53
| colt | d31efe70 | 2019-01-11 15:35:52
| berry| be2fcd50 | 2019-01-12 15:35:51
| colt | e754d050 | 2019-01-13 15:17:16
| colt | 9c87a7f0 | 2019-01-14 15:35:54
| colt | ucgtdes0 | 2019-01-15 12:30:59

    WITH dates(DATE) AS
      (SELECT DISTINCT Cast(check_in_date AS DATE),
                       user_id
       FROM check_ins),
         GROUPS AS
      (SELECT Row_number() OVER (
                                ORDER BY DATE) AS rn, DATE - (Row_number() OVER (ORDER BY DATE) * interval '1' DAY) AS grp, DATE, user_id
       FROM dates)
    SELECT Count(*) AS streak,
           user_id
    FROM GROUPS
    GROUP BY grp,
             user_id
    ORDER BY 1 DESC; 

以下是我运行上述代码时得到的结果：

 streak user_id
 --------------
 4      colt
 1      colt
 1      berry

它应该是什么。我还想为每个用户获得最长的连胜记录

 streak user_id
 --------------
 3      colt
 1      berry

---

首先，感谢fiddle脚本和示例数据

您没有使用正确的

行编号

来解决间隙和孤岛问题。它应该类似于下面的数据集查询。除此之外，要获得条纹最高的条纹，您需要在按组号分组后使用

DISTINCT On

（

grp

，在您的查询中，我称之为

seq

）

我希望您希望每天只看到用户数据的不同条目。我试图通过with子句中的细微变化来反映这一点

SELECT * FROM (  
WITH check_ins_dt AS
      ( SELECT DISTINCT check_in_date::DATE as check_in_date,

                       user_id
       FROM check_ins) 
SELECT DISTINCT ON (user_id) COUNT(*) AS streak,user_id

FROM (
     SELECT c.*,
            ROW_NUMBER() OVER(
                 ORDER BY check_in_date
            ) - ROW_NUMBER() OVER(
                 PARTITION BY user_id
                 ORDER BY check_in_date
            ) AS seq
     FROM check_ins_dt c
) s
GROUP BY user_id,
         seq
ORDER BY user_id,
COUNT(*) DESC ) q order
     by streak desc;

---

在Postgres中，您可以这样写：

select distinct on (user_id) user_id, count(distinct check_in_date::date) as num_days
from (select ci.*,
             dense_rank() over (partition by user_id order by check_in_date::date) as seq
      from check_ins ci
     ) ci
group by user_id, check_in_date::date - seq * interval '1 day'
order by user_id, num_days desc;

他是一把小提琴

---

这与您的方法遵循类似的逻辑，但您的查询似乎过于复杂。这确实使用了Postgres

distinct on

功能，这很方便避免额外的子查询。

mysql还是postgresql？用你正在使用的数据库解决你的问题。嘿，戈登！这看起来很好，很紧凑，而且很有效。唯一让我感到困惑的是按天数降序排列结果。@twobergs。为此，您需要一个子查询。感谢您的解决方案！