Question

如何获得上个月的所有星期一或星期二？我还没有看到任何关于它的例子。

Answer 1

您可以使用：

DECLARE @d DATE = GETDATE();

SELECT sub.prev_date
FROM (SELECT @d, MONTH(DATEADD(MM, -1, @d))) AS s(d,m)
CROSS APPLY (
    SELECT DATEADD(D, c-1, DATEADD(MM, -1, DATEADD(DD, 1 - DAY(d),d))) AS prev_date
    FROM (
    VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),
           (11),(12),(13),(14),(15),(16),(17),(18),(19),(20),
           (21),(22),(23),(24),(25),(26),(27),(28),(29),(30),(31))AS x(c)
    ) AS sub
WHERE MONTH(sub.prev_date) = s.m 
  AND DATENAME(dw,sub.prev_date) IN ('Monday','Tuesday');

的 LiveDemo

输出：

╔════════════╗ ║ prev_date ║ ╠════════════╣ ║ 2016-03-01 ║ ║ 2016-03-07 ║ ║ 2016-03-08 ║ ║ 2016-03-14 ║ ║ 2016-03-15 ║ ║ 2016-03-21 ║ ║ 2016-03-22 ║ ║ 2016-03-28 ║ ║ 2016-03-29 ║ ╚════════════╝

警告：

SQL Server语言应该English其他DATENAME不匹配。

您还可以与DATEPART weekday进行比较，但是您需要了解SET DATEFIRST设置。

修改

稍短一些：

DECLARE @d DATE = '2015-01-01'; SELECT sub.prev_date FROM (SELECT DATEADD(DD, c - DAY(@d),DATEADD(MM, -1, @d)) AS prev_date FROM (VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10), (11),(12),(13),(14),(15),(16),(17),(18),(19),(20), (21),(22),(23),(24),(25),(26),(27),(28),(29),(30),(31))AS x(c)) AS sub WHERE MONTH(sub.prev_date) = MONTH(DATEADD(MM, -1, @d)) AND DATENAME(dw,sub.prev_date) IN ('Monday','Tuesday');

的 LiveDemo

Answer 2

declare @table table
(
   ID integer identity,
   DateRange date,
   Remark nvarchar(10),
   MonWeek nvarchar(1),
   weekgrp integer
)

declare @from Date
declare @newfrom Date
declare @to Date
declare @min integer
declare @max integer

set @from ='2016-03-01'
set @to = '2016-03-31'
set @newfrom = '2016-03-01'

while @from <= @to
begin
  insert into @table (DateRange, Remark) Values (@from,DATENAME(dw,@from))

  set @from = DATEADD(dd,1,@from)
end

update @table
set MonWeek = 'Y'
where Remark = 'Monday'

select @min = MIN(ID), @max = MAX(ID) from @table
where MonWeek = 'Y'

--to calculate week group
while @min <= @max
begin
   Update @table
   set weekgrp = @min
   where ID between @min and @min + 7
   set @min = @min + 7
end

select * from @table
where Remark in ('Monday','Tuesday')

Answer 3

;WITH CTE (X)
 AS
 (
 SELECT DATEADD(MM,DATEDIFF(MM,0,GETDATE())-1,0)
 ),
 CTE2(N) AS
 (
 SELECT 0
 UNION ALL
 SELECT 1+N  FROM CTE2 WHERE N< (SELECT DATEDIFF(DD,DATEADD(MM,DATEDIFF(MM,0,GETDATE())-1,0),DATEADD(MM,1,DATEADD(MM,DATEDIFF(MM,0,GETDATE())-1,0))-1))
 )
 SELECT DATEADD(DD,N,X),DATENAME(DW,DATEADD(DD,N,X)) FROM CTE,CTE2 WHERE DATENAME(DW,DATEADD(DD,N,X)) IN ('Monday','Tuesday')

如何获得上个月的所有星期一或星期二

3 个答案: