SQL返回连续记录

一个简单的表格：

ForumPost
--------------
ID (int PK)
UserID (int FK)
Date (datetime)

我想要返回的是一个特定用户连续n天每天至少发表一篇文章的次数

例如：

User 15844 has posted at least 1 post a day for 30 consecutive days 10 times

---

我已经用linq/lambda标记了这个问题，还有一个解决方案也会很好。我知道我可以通过迭代所有用户记录来解决这个问题，但这很慢。

有一个简便的技巧，你可以使用

行号（）

来查找连续的条目，想象以下一组日期，它们的行号（从0开始）：

对于连续条目，如果从值中减去行数，则得到相同的结果。e、 g

Date        RowNumber   date - row_number
20130401    0           20130401
20130402    1           20130401
20130403    2           20130401
20130404    3           20130401
20130406    4           20130402
20130407    5           20130402

然后，您可以按

日期-行号

分组，以获得连续天数集（即前4条记录和最后2条记录）

要将此应用于您的示例，您可以使用：

WITH Posts AS
(   SELECT  FirstPost = DATEADD(DAY, 1 - ROW_NUMBER() OVER(PARTITION BY UserID ORDER BY [Date]), [Date]),
            UserID,
            Date
    FROM    (   SELECT  DISTINCT UserID, [Date] = CAST(Date AS [Date])
                FROM    ForumPost
            ) fp
), Posts2 AS
(   SELECT  FirstPost, 
            UserID, 
            Days = COUNT(*), 
            LastDate = MAX(Date)
    FROM    Posts
    GROUP BY FirstPost, UserID
)
SELECT  UserID, ConsecutiveDates = MAX(Days)
FROM    Posts2
GROUP BY UserID;

编辑

我不认为上面的回答完全正确，这将给出一个用户发布的次数或连续n天的次数：

WITH Posts AS
(   SELECT  FirstPost = DATEADD(DAY, 1 - ROW_NUMBER() OVER(PARTITION BY UserID ORDER BY [Date]), [Date]),
            UserID,
            Date
    FROM    (   SELECT  DISTINCT UserID, [Date] = CAST(Date AS [Date])
                FROM    ForumPost
            ) fp
), Posts2 AS
(   SELECT  FirstPost, 
            UserID, 
            Days = COUNT(*), 
            FirstDate = MIN(Date), 
            LastDate = MAX(Date)
    FROM    Posts
    GROUP BY FirstPost, UserID
)
SELECT  UserID, [Times Over N Days] = COUNT(*)
FROM    Posts2
WHERE   Days >= 30
GROUP BY UserID;

---

我认为，您的特定应用程序使这个过程非常简单。如果在“n”天间隔内有“n”个不同的日期，则这些“n”个不同的日期必须是连续的

滚动至底部，查看只需要通用表表达式并更改为PostgreSQL的通用解决方案。（开玩笑。我是用PostgreSQL实现的，因为我时间不够。）

现在，让我们看看这个查询的输出。为了简单起见，我考虑的是5天的间隔，而不是30天的间隔

select userid, count(distinct post_date) distinct_dates
from forumpost
where post_date between '2013-01-15' and '2013-01-19'
group by userid;

USERID  DISTINCT_DATES  
1       5
2       5
3       1

对于符合条件的用户，在5天的时间间隔内，不同日期的数量必须是5，对吗？所以我们只需要在HAVING子句中添加这个逻辑

select userid, count(distinct post_date) distinct_dates
from forumpost
where post_date between '2013-01-15' and '2013-01-19'
group by userid
having count(distinct post_date) = 5;

USERID  DISTINCT_DATES  
1       5
2       5

---

更通用的解决方案

如果你从2013-01-01到2013-01-31每天都发帖，那么你已经连续发帖30天2次了，这样说是没有意义的。相反，我希望时钟能在2013年1月31日重新开始。我为在PostgreSQL中实现而道歉；稍后我将尝试在T-SQL中实现

with first_posts as (
  select userid, min(post_date) first_post_date
  from forumpost
  group by userid
), 
period_intervals as (
  select userid, first_post_date period_start, 
         (first_post_date + interval '4' day)::date period_end
  from first_posts
), user_specific_intervals as (
  select 
    userid, 
    (period_start + (n || ' days')::interval)::date as period_start, 
    (period_end + (n || ' days')::interval)::date as period_end 
  from period_intervals, generate_series(0, 30, 5) n
)
select userid, period_start, period_end, 
       (select count(distinct post_date) 
        from forumpost
        where forumpost.post_date between period_start and period_end
          and userid = forumpost.userid) distinct_dates
from user_specific_intervals
order by userid, period_start;

---

您使用的是哪种数据库管理系统？博士后？Oracle？对日期范围为30天前的所有帖子使用子查询，按日期和计数分组。。检查是否30？相关，可能重复：@FuzzyButton-如果他们一天发布30次，而29天没有发布，那么总数仍然是30。这取决于“连续30天10次”的含义。如果你在2013-01-01和2013-01-30之间每天都发帖，那就是连续30天，1次。如果你在2013年1月31日再次发帖，这是连续30天2次吗？

select userid, count(distinct post_date) distinct_dates
from forumpost
where post_date between '2013-01-15' and '2013-01-19'
group by userid
having count(distinct post_date) = 5;

USERID  DISTINCT_DATES  
1       5
2       5

with first_posts as (
  select userid, min(post_date) first_post_date
  from forumpost
  group by userid
), 
period_intervals as (
  select userid, first_post_date period_start, 
         (first_post_date + interval '4' day)::date period_end
  from first_posts
), user_specific_intervals as (
  select 
    userid, 
    (period_start + (n || ' days')::interval)::date as period_start, 
    (period_end + (n || ' days')::interval)::date as period_end 
  from period_intervals, generate_series(0, 30, 5) n
)
select userid, period_start, period_end, 
       (select count(distinct post_date) 
        from forumpost
        where forumpost.post_date between period_start and period_end
          and userid = forumpost.userid) distinct_dates
from user_specific_intervals
order by userid, period_start;