Question

一个简单的表格：

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

我希望返回特定用户连续几天至少每天发布一次 n 的帖子。

示例：

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

我用linq / lambda标记了这个问题以及解决方案也会很棒。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。

Answer 1

使用ROW_NUMBER()可以使用一个方便的技巧来查找连续的条目，想象下面的日期集，以及它们的row_number（从0开始）：

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

对于连续条目，如果从值中减去row_number，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date - row_number分组以获取连续日数（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;

<强> Example on SQL Fiddle (simple with just most consecutive days per user)

<强> Further example to show how to get all consecutive periods

编辑

我不认为上述内容已经回答了这个问题，这将给出用户发布或连续n天发布的次数：

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;

<强> Example on SQL Fiddle

Answer 2

我认为你的特定应用程序非常简单。如果你有＆＃39; n＆＃39;在“＆n” - 天的间隔中的不同日期，那些＆nbsp;＆＃39; n＆＃39;不同的日期必须是连续的。

滚动到底部以获得一般解决方案，该解决方案仅需要公用表表达式并更改为PostgreSQL。（开玩笑。我在PostgreSQL中实现，因为我的时间不够。）

create table ForumPost (
  ID integer primary key,
  UserID integer not null,
  post_date date not null
);

insert into forumpost values
(1, 1, '2013-01-15'),
(2, 1, '2013-01-16'),
(3, 1, '2013-01-17'),
(4, 1, '2013-01-18'),
(5, 1, '2013-01-19'),
(6, 1, '2013-01-20'),
(7, 1, '2013-01-21'),

(11, 2, '2013-01-15'),
(12, 2, '2013-01-16'),
(13, 2, '2013-01-17'),
(16, 2, '2013-01-17'),
(14, 2, '2013-01-18'),
(15, 2, '2013-01-19'),

(21, 3, '2013-01-17'),
(22, 3, '2013-01-17'),
(23, 3, '2013-01-17'),
(24, 3, '2013-01-17'),
(25, 3, '2013-01-17'),
(26, 3, '2013-01-17'),
(27, 3, '2013-01-17');

现在，让我们看看这个查询的输出。为简洁起见，我每隔5天检查一次，而不是每隔30天。

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，该5天间隔内的不同日期数必须为5，对吗？所以我们只需要将该逻辑添加到HAVING子句中。

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

如果您每天从2013-01-01发布到2013-01-31，那么您已经连续30天发布了2次，这没有任何意义。相反，我希望时钟能够在2013-01-31重新开始。我对在PostgreSQL中实现表示歉意;我稍后会尝试在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;

SQL返回连续记录

2 个答案: