Question

我需要sql查询，它通过传递两个日期在所有日期之间给出结果。例如，假设我发送了01/01/2015和10/01/2015

我需要传递日期之间的日期

结果应为：

我怎样才能做到这一点？

Answer 1

您可以使用recursive CTE：

DECLARE @startDate DATETIME = '2015-01-01',
    @endDate DATETIME = '2015-01-10'

;WITH dates AS(
 SELECT @startDate AS date
 UNION ALL
 SELECT DATEADD(dd,1,date) 
 FROM dates
 WHERE date<@endDate
)
SELECT * FROM dates

Answer 2

使用Numbers table生成日期。

DECLARE @StartDate DATETIME = convert(date,'01/01/2015',103) -- First Calendar date to include in table
DECLARE @EndDate DATETIME = convert(date,'10/01/2015',103) --Last calendar date to include in the table


;WITH   E00(N) AS (SELECT 1 UNION ALL SELECT 1),
        E02(N) AS (SELECT 1 FROM E00 a, E00 b),
        E04(N) AS (SELECT 1 FROM E02 a, E02 b),
        E08(N) AS (SELECT 1 FROM E04 a, E04 b),
        E16(N) AS (SELECT 1 FROM E08 a, E08 b),
        E32(N) AS (SELECT 1 FROM E16 a, E16 b),
     cteTally(N)
     AS (SELECT Row_number()OVER ( ORDER BY N)
         FROM   E32),
     CalendarBase
     AS (SELECT CalendarDate = Dateadd(day, n - 1, @StartDate)
         FROM   cteTally
         WHERE  N <= Datediff(day, @StartDate, @EndDate + 1))
SELECT convert(varchar(30),CalendarDate ,103) as  CalendarDate
FROM   CalendarBase

从此link

引用

Answer 3

试试这个：

declare @strt date;
declare @end date;

select @strt = '20150101';
select @end = '20150110';

with dates as
(
select dt = dateadd(dd, 0, @strt)
where dateadd(dd, 1, @strt) <= @end
union all
select dateadd(dd, 1, dt)
from dates
where dateadd(dd, 1, dt) <= @end
)
select * from dates

Answer 4

@Saurabh Dhakate

Use the below code:

declare @date1 date ='2015-01-10'
declare @date2 date  ='2015-01-01'
declare @lv_table table (datepart1 date)
declare @count int 
declare @count1 int = 1
set @count=(select day(@date1) )
print @count
while @count1<=@count
begin
insert into @lv_table
select dateadd(day,@count1,@date2)
set @count1=@count1+1
print @count1
end
select * from  @lv_table

SQL服务器查找两个选定日期之间的日期

4 个答案: