2

我有以下。

Name    Date
A   2011-01-01 01:00:00.000
A   2011-02-01 02:00:00.000
A   2011-03-01 03:00:00.000
B   2011-04-01 04:00:00.000
A   2011-05-01 07:00:00.000

所需的输出是

Name       StartDate                        EndDate
-------------------------------------------------------------------
A          2011-01-01 01:00:00.000         2011-04-01 04:00:00.000    
B          2011-04-01 04:00:00.000         2011-05-01 07:00:00.000    
A          2011-05-01 07:00:00.000         NULL

如何在基于集合的方法中使用 TSQL 实现相同的目的。

DDL如下

DECLARE @t TABLE(PersonName VARCHAR(32), [Date] DATETIME) 
INSERT INTO @t VALUES('A', '2011-01-01 01:00:00') 
INSERT INTO @t VALUES('A', '2011-01-02 02:00:00') 
INSERT INTO @t VALUES('A', '2011-01-03 03:00:00') 
INSERT INTO @t VALUES('B', '2011-01-04 04:00:00') 
INSERT INTO @t VALUES('A', '2011-01-05 07:00:00')

Select * from @t
4

5 回答 5

7 
;WITH cte1
     AS (SELECT *,
                ROW_NUMBER() OVER (ORDER BY Date) -
                ROW_NUMBER() OVER (PARTITION BY PersonName
                ORDER BY Date) AS G
         FROM   @t),
     cte2
     AS (SELECT PersonName,
                MIN([Date]) StartDate,
                ROW_NUMBER() OVER (ORDER BY  MIN([Date])) AS rn
         FROM   cte1
         GROUP  BY PersonName,
                   G)
SELECT a.PersonName,
       a.StartDate,
       b.StartDate AS EndDate
FROM   cte2 a
       LEFT JOIN cte2 b
         ON a.rn + 1 = b.rn

因为 CTE 的结果通常不会具体化，但是如果您自己实现中间结果，您可能会发现您会获得更好的性能，如下所示。

DECLARE @t2 TABLE (
  rn         INT IDENTITY(1, 1) PRIMARY KEY,
  PersonName VARCHAR(32),
  StartDate  DATETIME );

INSERT INTO @t2
SELECT PersonName,
       MIN([Date]) StartDate
FROM   (SELECT *,
               ROW_NUMBER() OVER (ORDER BY Date) -
               ROW_NUMBER() OVER (PARTITION BY PersonName
               ORDER BY Date) AS G
        FROM   @t) t
GROUP  BY PersonName,
          G
ORDER  BY StartDate

SELECT a.PersonName,
       a.StartDate,
       b.StartDate AS EndDate
FROM   @t2 a
       LEFT JOIN @t2 b
         ON a.rn + 1 = b.rn
于 2011-01-10T04:56:05.383 回答
1 
SELECT
  PersonName,
  StartDate = MIN(Date),
  EndDate
FROM (
  SELECT
    PersonName,
    Date,
    EndDate = (
      /* get the earliest date after current date
         associated with a different person */
      SELECT MIN(t1.Date)
      FROM @t AS t1
      WHERE t1.Date > t.Date
        AND t1.PersonName <> t.PersonName
    )
  FROM @t AS t
) s
GROUP BY PersonName, EndDate
ORDER BY 2

基本上，对于每一个Date我们找到它之后最近的日期，它都与不同的PersonName. 这给了我们EndDate，现在为我们区分同一个人的连续日期组。

现在我们只需要按PersonName&对数据进行分组，EndDate并将每个组中的最小值Date作为StartDate. 是的，当然，按 对数据进行排序StartDate。

于 2011-01-10T09:55:22.733 回答
0

cte 的另一个答案是一个很好的答案。另一种选择是在任何情况下都遍历集合。它不是基于集合的，但它是另一种方法。

您将需要迭代到 A. 为与其事务对应的每条记录分配一个唯一的 id，或者 B. 以实际获取您的输出。

TSQL 不适合迭代记录，特别是如果你有很多，所以我会推荐一些其他的方法，一个小的 .net 程序或者更擅长迭代的东西。

于 2011-01-10T04:44:38.717 回答
0

获取行号，以便您知道上一条记录在哪里。然后，取一个记录和它之后的下一个记录。当状态改变时，我们有一个候选行。

select 
  state, 
  min(start_timestamp),
  max(end_timestamp)

from
(
    select
        first.state, 
        first.timestamp_ as start_timestamp,
        second.timestamp_ as end_timestamp

        from
        (
            select
                *, row_number() over (order by timestamp_) as id
            from test
        ) as first

        left outer join
        (
            select
                *, row_number() over (order by timestamp_) as id
            from test
        ) as second
        on 
            first.id = second.id - 1 
            and first.state != second.state
) as agg
group by state
    having max(end_timestamp) is not null 

union

-- last row wont have a ending row
--(select state, timestamp_, null from test order by timestamp_ desc limit 1)
    -- I think it something like this for sql server
     (select top state, timestamp_, null from test order by timestamp_ desc)

order by 2
;

已使用 PostgreSQL 测试，但也应与 SQL Server 一起使用

于 2011-01-10T06:38:22.083 回答
0

使用一些间隙和岛屿理论有一个非常快速的方法：

WITH CTE as (SELECT PersonName, [Date]
                   , Row_Number() over (ORDER BY [Date])
                     - Row_Number() over (ORDER BY PersonName, [Date]) as Island
             FROM @t)

Select PersonName, Min([Date]), Max([Date])
from CTE
GROUP BY Island, PersonName
ORDER BY Min([Date])
于 2018-05-18T14:44:36.730 回答