Sql server 如何在SQL Server中查找日期间隔?
我有一张桌子Sql server 如何在SQL Server中查找日期间隔?,sql-server,tsql,Sql Server,Tsql,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 20
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
输出将是
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,2014-07-04之后的下一个工作日为“2014-07-08”,此后2014-07-05之后的产量为2014-07-08。
2014-07-10之后的下一个工作日为“2014-07-11”,此后2014-07-10之后的产量为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
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
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
检查此查询,希望它能帮助您: 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)
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
所需输出:
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
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
检查此查询,希望它能帮助您: 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)
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
所需输出:
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
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
2014-07-04
的下一个工作日不应该是2014-07-08
?2014-07-04
的下一个工作日不应该是2014-07-08
?