Question

我正在根据员工工作的时间块做一些报告。在某些情况下，数据包含两个单独的记录，用于确切的单个时间块。

这是该表的基本版本和一些示例记录：

EmployeeID
StartTime
EndTime

数据：

EmpID      Start         End
----------------------------
#1001   10:00 AM    12:00 PM
#1001    4:00 PM     5:30 PM
#1001    5:30 PM     8:00 PM

在该示例中，最后两个记录在时间上是连续的。我想写一个组合任何相邻记录的查询，所以结果集是这样的：

EmpID      Start         End
----------------------------
#1001   10:00 AM    12:00 PM
#1001    4:00 PM     8:00 PM

理想情况下，它还应该能够处理2个以上的相邻记录，但这不是必需的。

Answer 1

本文为您的问题提供了很多可能的解决方案

http://www.sqlmag.com/blog/puzzled-by-t-sql-blog-15/tsql/solutions-to-packing-date-and-time-intervals-puzzle-136851

这个似乎是最直接的：

WITH StartTimes AS
(
  SELECT DISTINCT username, starttime
  FROM dbo.Sessions AS S1
  WHERE NOT EXISTS
    (SELECT * FROM dbo.Sessions AS S2
     WHERE S2.username = S1.username
       AND S2.starttime < S1.starttime
       AND S2.endtime >= S1.starttime)
),
EndTimes AS
(
  SELECT DISTINCT username, endtime
  FROM dbo.Sessions AS S1
  WHERE NOT EXISTS
    (SELECT * FROM dbo.Sessions AS S2
     WHERE S2.username = S1.username
       AND S2.endtime > S1.endtime
       AND S2.starttime <= S1.endtime)
)
SELECT username, starttime,
  (SELECT MIN(endtime) FROM EndTimes AS E
   WHERE E.username = S.username
     AND endtime >= starttime) AS endtime
FROM StartTimes AS S;

Answer 2

如果这是严格关于相邻行（不重叠的行），您可以尝试以下方法：

取消隐藏时间戳。
只留下那些没有重复的内容。
将其余部分转回，将Start与End后面的WITH unpivoted AS ( SELECT EmpID, event, dtime, count = COUNT(*) OVER (PARTITION BY EmpID, dtime) FROM atable UNPIVOT ( dtime FOR event IN (StartTime, EndTime) ) u ) , filtered AS ( SELECT EmpID, event, dtime, rowno = ROW_NUMBER() OVER (PARTITION BY EmpID, event ORDER BY dtime) FROM unpivoted WHERE count = 1 ) , pivoted AS ( SELECT EmpID, StartTime, EndTime FROM filtered PIVOT ( MAX(dtime) FOR event IN (StartTime, EndTime) ) p ) SELECT * FROM pivoted ;联系起来。

或者，在Transact-SQL中，类似这样：

{{1}}

此查询有at SQL Fiddle的演示。

Answer 3

我已经改变了一些名称和类型以使示例变小但是这样做并且应该非常快并且没有记录限制数量：

with cte as (
  select 
    x1.id
    ,x1.t1
    ,x1.t2
    ,case when x2.t1 is null then 1 else 0 end as bef
    ,case when x3.t1 is null then 1 else 0 end as aft
  from x x1
  left join x x2 on x1.id=x2.id and x1.t1=x2.t2
  left join x x3 on x1.id=x3.id and x1.t2=x3.t1
  where x2.id is null
  or    x3.id is null
)

select 
  cteo.id
  ,cteo.t1
  ,isnull(z.t2,cteo.t2) as t2

from cte cteo
outer apply (select top 1 * 
             from cte ctei 
             where cteo.id=ctei.id and cteo.aft=0 and ctei.t1>cteo.t1
             order by t1) z
where cteo.bef=1

和它的小提琴：http://sqlfiddle.com/#!3/ad737/12/0

Answer 4

具有内联用户定义函数和CTE的选项

CREATE FUNCTION dbo.Overlap
 (
  @availStart datetime,
  @availEnd datetime,
  @availStart2 datetime,
  @availEnd2 datetime
  )
RETURNS TABLE
RETURN
  SELECT CASE WHEN @availStart > @availEnd2 OR @availEnd < @availStart2
              THEN @availStart ELSE
                               CASE WHEN @availStart > @availStart2 THEN @availStart2 ELSE @availStart END
                               END AS availStart,
         CASE WHEN @availStart > @availEnd2 OR @availEnd < @availStart2
              THEN @availEnd ELSE
                             CASE WHEN @availEnd > @availEnd2 THEN @availEnd ELSE @availEnd2 END
                             END AS availEnd

;WITH cte AS
 (
  SELECT EmpID, Start, [End], ROW_NUMBER() OVER (PARTITION BY EmpID ORDER BY Start) AS Id
  FROM dbo.TableName
  ), cte2 AS
 (
  SELECT Id, EmpID, Start, [End]
  FROM cte
  WHERE Id = 1
  UNION ALL
  SELECT c.Id, c.EmpID, o.availStart, o.availEnd
  FROM cte c JOIN cte2 ct ON c.Id = ct.Id + 1
             CROSS APPLY dbo.Overlap(c.Start, c.[End], ct.Start, ct.[End]) AS o
  )
  SELECT EmpID, Start, MAX([End])
  FROM cte2
  GROUP BY EmpID, Start

SQLFiddle上的演示

Answer 5

累积金额的CTE：

DECLARE @t TABLE(EmpId INT, Start TIME, Finish TIME)
INSERT INTO @t (EmpId, Start, Finish)
VALUES
    (1001, '10:00 AM', '12:00 PM'),
    (1001, '4:00 PM', '5:30 PM'),
    (1001, '5:30 PM', '8:00 PM')

;WITH rowind AS (
    SELECT EmpId, Start, Finish,
        -- IIF returns 1 for each row that should generate a new row in the final result
        IIF(Start = LAG(Finish, 1) OVER(PARTITION BY EmpId ORDER BY Start), 0, 1) newrow
    FROM @t),
    groups AS (
    SELECT EmpId, Start, Finish,
        -- Cumulative sum
        SUM(newrow) OVER(PARTITION BY EmpId ORDER BY Start) csum
    FROM rowind)

SELECT
    EmpId,
    MIN(Start) Start,
    MAX(Finish) Finish
FROM groups
GROUP BY EmpId, csum

合并SQL中的相邻行？

5 个答案: