Question

我需要结果中的所有日期排除星期日：

输入：开始日期= 2016-07-01和结束日期= 2016-07-31

S_No    Dates
1   01-07-2016
2   02-07-2016
3   04-07-2016
4   05-07-2016
5   06-07-2016
6   07-07-2016
7   08-07-2016
8   09-07-2016
9   11-07-2016
10  12-07-2016
11  13-07-2016
12  14-07-2016
13  15-07-2016
14  16-07-2016
15  18-07-2016
16  19-07-2016
17  20-07-2016
18  21-07-2016
19  22-07-2016
20  23-07-2016
21  25-07-2016
22  26-07-2016
23  27-07-2016
24  28-07-2016
25  29-07-2016
26  30-07-2016

我需要这样的结果。如果我输入输入：Start Date = 2016-06-01 and End Date = 2016-06-30。所有日期都应该在2016年6月排除星期日。

请帮帮我？

Answer 1

您可以使用给定的查询

此dw函数中的datepart()表示＆＃34;周日＆＃34;。你想要排除星期日，所以在where子句的条件中写1。

查询：

select S_No, Dates
from table_name 
where datepart(dw,Dates) <>1 and Dates >='2016-06-01' and Dates<='2016-06-30'

因此，根据此查询，您将获得2016-06-01和2016-06-30之间的所有日期，不包括星期日。

Answer 2

如果您想要一个返回不包括星期日的日期的查询，请尝试以下方法：

declare @dateStart date
declare @dateEnd date

set datefirst 7 -- Setting Sunday as firstday of the week

set @dateStart = cast('2016-06-01' as date) -- Maybe as paramters?
set @dateEnd = cast('2016-06-30' as date)



;WITH T AS (SELECT @dateStart as d union all
            SELECT DATEADD(DAY,1,d) from T where d<@dateEnd)
select *
  from T
where datepart(WEEKDAY,d) <> 1

<强>输出

Answer 3

创建一个整数表：

SELECT TOP 1000000
        IDENTITY(INT,1,1) as N
   INTO dbo.NumbersTable
   FROM Master.dbo.SysColumns

创建日期范围表

DECLARE @DaysFromStart int
DECLARE @StartDate datetime
SET @StartDate = '10/01/2008'

SET @DaysFromStart  = (SELECT (DATEDIFF(dd,@StartDate,GETDATE()) + 1))

CREATE TABLE [dbo].[TableOfDates](
    [fld_date] [datetime] NOT NULL,
 CONSTRAINT [PK_TableOfDates] PRIMARY KEY CLUSTERED 
(
    [fld_date] ASC
)WITH FILLFACTOR = 99 ON [PRIMARY]
) ON [PRIMARY]


INSERT INTO
     dbo.TableOfDates
SELECT 
      DATEADD(dd,nums.n - @DaysFromStart ,CAST(FLOOR(CAST(GETDATE() as FLOAT)) as DateTime)) as FLD_Date
FROM dbo.NumbersTable nums

SELECT MIN(FLD_Date) FROM dbo.TableOfDates
SELECT MAX(FLD_Date) FROM dbo.TableOfDates

您要使用的最终查询：

SELECT FLD_Date FROM TableOfDates (NOLOCK) WHERE FLD_Date >= '2016-07-01' AND FLD_Date <= '2016-07-31' AND DATENAME(dw,FLD_Date) <> 'Sunday'

参考：Stack overflow

Answer 4

检查一下

WITH mycte AS
 (
   SELECT CAST('2016-06-01' AS DATETIME) DateValue
   UNION ALL
   SELECT  DateValue + 1
   FROM    mycte   
   WHERE   DateValue + 1 < '2016-07-01'
 )

 SELECT  DateValue
 FROM    mycte
 WHERE ((DATEPART(dw, DateValue) + @@DATEFIRST) % 7) NOT IN (1)
 OPTION (MAXRECURSION 0);

输出将是......

DateValue
2016-06-01 00:00:00.000
2016-06-02 00:00:00.000
2016-06-03 00:00:00.000
2016-06-04 00:00:00.000
2016-06-06 00:00:00.000
2016-06-07 00:00:00.000
2016-06-08 00:00:00.000
2016-06-09 00:00:00.000
2016-06-10 00:00:00.000
2016-06-11 00:00:00.000
2016-06-13 00:00:00.000
2016-06-14 00:00:00.000
2016-06-15 00:00:00.000
2016-06-16 00:00:00.000
2016-06-17 00:00:00.000
2016-06-18 00:00:00.000
2016-06-20 00:00:00.000
2016-06-21 00:00:00.000
2016-06-22 00:00:00.000
2016-06-23 00:00:00.000
2016-06-24 00:00:00.000
2016-06-25 00:00:00.000
2016-06-27 00:00:00.000
2016-06-28 00:00:00.000
2016-06-29 00:00:00.000
2016-06-30 00:00:00.000

Answer 5

您的问题的解决方案如下：

DECLARE @START_DATE DATETIME='2:57 PM 7/01/2016',@END_DATE DATETIME='2:57 PM 7/26/2016'


WHILE(@START_DATE<=@END_DATE)
BEGIN
IF(DATENAME(WEEKDAY,@START_DATE)<>'Sunday')
BEGIN
   PRINT @START_DATE --Modify code according to your need.
END
SELECT @START_DATE=DATEADD(D,1,@START_DATE)
END

结果：

Jul  1 2016  2:57PM
Jul  2 2016  2:57PM
Jul  4 2016  2:57PM
Jul  5 2016  2:57PM
Jul  6 2016  2:57PM
Jul  7 2016  2:57PM
Jul  8 2016  2:57PM
Jul  9 2016  2:57PM
Jul 11 2016  2:57PM
Jul 12 2016  2:57PM
Jul 13 2016  2:57PM
Jul 14 2016  2:57PM
Jul 15 2016  2:57PM
Jul 16 2016  2:57PM
Jul 18 2016  2:57PM
Jul 19 2016  2:57PM
Jul 20 2016  2:57PM
Jul 21 2016  2:57PM
Jul 22 2016  2:57PM
Jul 23 2016  2:57PM
Jul 25 2016  2:57PM
Jul 26 2016  2:57PM

Answer 6

SELECT * FROM tableName 在2016-06-01 03：13：45.040'''2016-06-30 03：13：45.040'和DATEPART（dw，日期）＆lt;＆gt; 1

之间的日期

我需要结果中的所有日期排除星期日

6 个答案: