如何按日期范围对记录进行分组

Question

如何在sql中按日期范围对记录进行分组？

考虑一下这个表结构。

 Key       ID       VISITDATE
 1         1        2011-01-07
 2         1        2011-01-09
 3         2        2011-01-10
 4         1        2011-01-12
 5         3        2011-01-12
 6         1        2011-01-18
 7         2        2011-01-21
 9         1        2011-02-28
 10        2        2011-03-21
 11        1        2011-01-06
 12        1        2011-02-29

我需要获取出现次数。如果您在 10 天内对同一个 ID 进行了两次访问，则应仅计算一次访问。 因此，对于给定 id 1 = 3 (01/06,01/18,02/28) 的出现次数的示例

非常感谢递归 CTE 查询。

Answer 1

我知道您要求递归 CTE，但没有 SQL Server 2012 中的新窗口增强功能（FIRST_VALUE() 等）。我认为，当您必须跟踪仅前一行，但同时还有较早的行。这是我相信可以实现您想要的光标版本：

DECLARE @f TABLE([Key] INT, ID INT, VISITDATE DATE);

INSERT @f VALUES
  (1 ,1,'2011-01-07'), (2 ,1,'2011-01-09'), (3 ,2,'2011-01-10'), (4 ,1,'2011-01-12'),
  (5 ,3,'2011-01-12'), (6 ,1,'2011-01-18'), (7 ,2,'2011-01-21'), (9 ,1,'2011-02-28'), 
  (10,2,'2011-03-21'), (11,1,'2011-01-06'), (12,1,'2011-03-01');

DECLARE @ID INT, @dt DATE;

DECLARE @result TABLE(ID INT, FirstDate DATE, VisitCount INT);

DECLARE c CURSOR LOCAL STATIC READ_ONLY FORWARD_ONLY
    FOR SELECT ID, VISITDATE FROM @f ORDER BY ID, VISITDATE;

OPEN c;

FETCH NEXT FROM c INTO @ID, @dt;

WHILE @@FETCH_STATUS = 0
BEGIN
    IF NOT EXISTS 
    (
        SELECT 1 FROM @result WHERE ID = @ID
            AND DATEDIFF(DAY, FirstDate, @dt) <= 10 -- maybe < 10?
    )
    INSERT @result SELECT @ID, @dt, 1;

    FETCH NEXT FROM c INTO @ID, @dt;
END

SELECT ID, FirstDate FROM @result;

SELECT ID, VisitCount = COUNT(*) FROM @result GROUP BY ID;

CLOSE c;
DEALLOCATE c;

结果：

ID          FirstDate
----------- ----------
1           2011-01-06
1           2011-01-18
1           2011-02-28
2           2011-01-10
2           2011-01-21
2           2011-03-21
3           2011-01-12


ID          VisitCount
----------- -----------
1           3
2           3
3           1

是的，我知道您总是被警告不要与陌生人交谈并远离光标，但在某些情况下，它们是最直接的解决方案（并且有时比基于集合的解决方案的多次扫描执行得更好可能会发生）。

Answer 2

您没有说明如何定义日期范围

如果从今天开始，你可以这样做：

SELECT     id, COUNT(DISTINCT DATEDIFF(dd, visitdate, GETDATE()) / 10) AS Expr1
FROM         test
GROUP BY id

如果您想从数据的最新日期开始范围，您可以这样做

DECLARE @maxdate as datetime
SET @maxdate = (select MAX(visitdate) from test)

SELECT     id, COUNT(DISTINCT DATEDIFF(dd, visitdate, @maxdate) / 10) 
FROM         test
GROUP BY id

这样的查询可以解决@Dems 指出的情况

SELECT id, count(DISTINCT visitdate)
FROM (
    SELECT   id, visitdate,
        (SELECT MAX(visitdate) 
        FROM test AS t 
        WHERE t.id = test.id AND t.visitdate<test.visitdate) AS prev_date
    FROM     test ) as temp
WHERE DATEDIFF(dd, prev_date, visitdate) > 10 OR prev_date IS NULL
GROUP BY id

上述查询不计算具有相同 id 且存在另一条日期小于十天的记录的记录

如果您想从每个 id 的最小日期开始拆分日历，您可以执行以下操作

SELECT     t.id, COUNT(DISTINCT DATEDIFF(dd, mindate, visitdate) / 10) 
FROM         test as t JOIN 
  (SELECT id, MIN(visitdate) AS mindate FROM test GROUP BY id) as mindates
    ON t.id = mindates.id
GROUP BY t.id

Answer 3

创建一个标量函数，为给定范围内的所有日期返回相同的日期，然后按该日期分组。