Question

我有一张桌子

DATE    LEAVEDAYS   NXTWRKDAYS
2014-07-01  No  NULL
2014-07-02  No  NULL
2014-07-03  No  NULL
2014-07-04  No  NULL
2014-07-05  Yes NULL
2014-07-06  Yes NULL
2014-07-07  Yes NULL
2014-07-08  No  NULL
2014-07-09  Yes NULL
2014-07-10  Yes NULL
2014-07-11  No  NULL

输出

DATE        LEAVEDAYS   NXTWRKDAYS
2014-07-01  No      2014-07-02
2014-07-02  No      2014-07-03
2014-07-03  No      2014-07-04
2014-07-04  No      2014-07-05
2014-07-05  Yes     2014-07-08
2014-07-06  Yes     2014-07-08
2014-07-07  Yes     2014-07-08
2014-07-08  No      2014-07-09 
2014-07-09  Yes     2014-07-11
2014-07-10  Yes     2014-07-11
2014-07-11  No      2014-07-12

可以看出，在2014-07-04之后，下一个工作日是＆＃39; 2014-07-08＆＃39;从此以后，2014-07-05之后的产量是2014-07-08。 2014-07-10之后，下一个工作日是2014-07-11＆＃39;从此以后，2014-07-10之后的产出是2014-07-11

脚本

declare @t table([DATE] DATE,LEAVEDAYS Varchar(10),NXTWRKDAYS DATE)
insert into @t Values
('2014-07-01','No',NULL),
('2014-07-02','No',NULL),
('2014-07-03','No',NULL),
('2014-07-04','No',NULL),
('2014-07-05','Yes',NULL),
('2014-07-06','Yes',NULL),
('2014-07-07','Yes',NULL),
('2014-07-08','No',NULL),
('2014-07-09','Yes',NULL),
('2014-07-10','Yes',NULL),
('2014-07-11','No',NULL)

select *
from @t

Answer 1

检查此查询，希望它可以帮助您：

您的创建表和插入脚本：

declare @t table([DATE] DATE,LEAVEDAYS Varchar(10),NXTWRKDAYS DATE)
insert into @t Values
('2014-07-01','No',NULL),
('2014-07-02','No',NULL),
('2014-07-03','No',NULL),
('2014-07-04','No',NULL),
('2014-07-05','Yes',NULL),
('2014-07-06','Yes',NULL),
('2014-07-07','Yes',NULL),
('2014-07-08','No',NULL),
('2014-07-09','Yes',NULL),
('2014-07-10','Yes',NULL),
('2014-07-11','No',NULL)

必填项：

SELECT 
     t.date AS Date,
     t.leavedays AS LeaveDays, 
     CASE WHEN t.LEAVEDAYS = 'No' THEN DATEADD(day,1, t.DATE) 
          WHEN t.LEAVEDAYS = 'Yes' THEN (Select top 1 t1.date from @t t1 WHERE t1.date > t.DATE and t1.LEAVEDAYS = 'No' ORDER BY t1.date) 
          ELSE null END AS NxtWorkingDay
FROM @t t

Answer 2

您可以使用APPLY和TOP：

获得所需的结果

SELECT
    t.[DATE],
    t.LEAVEDAYS,
    NXTWRKDAYS = x.[DATE]
FROM @t t
OUTER APPLY(
    SELECT TOP 1 [DATE]
    FROM @t
    WHERE 
        [DATE] > t.[DATE]
        AND LEAVEDAYS = CASE WHEN t.LEAVEDAYS = 'YES' THEN 'NO' ELSE LEAVEDAYS END          
) x

如何在SQL Server中查找日期中的空白？

2 个答案: