Get All Dates of Given Month and Year in SQL Serve

问题:

I want to get all dates by declaring month and year in SQL server.

Can anyone please share easy lines of SQL code to get it.

For example:

DECLARE @month AS INT = 5
DECLARE @Year AS INT = 2016
SELECT * from Something

I have tried below things,

DECLARE @month TINYINT=5

;WITH CTE_Days AS (
    SELECT DATEADD(
               MONTH,
               @month,
               DATEADD(
                   MONTH,
                   -MONTH(GETDATE()),
                   DATEADD(
                       DAY,
                       -DAY(GETDATE()) + 1,
                       CAST(FLOOR(CAST(GETDATE() AS FLOAT)) AS DATETIME)
                   )
               )
           ) Dates
    UNION ALL
    SELECT DATEADD(DAY, 1, Dates)
    FROM   CTE_Days
    WHERE  Dates < DATEADD(
               DAY,
               -1,
               DATEADD(
                   MONTH,
                   1,
                   DATEADD(
                       MONTH,
                       @month,
                       DATEADD(
                           MONTH,
                           -MONTH(GETDATE()),
                           DATEADD(
                               DAY,
                               -DAY(GETDATE()) + 1,
                               CAST(FLOOR(CAST(GETDATE() AS FLOAT)) AS DATETIME)
                           )
                       )
                   )
               )
           )
)
SELECT Dates
FROM   CTE_Days

But I am looking for easy solution with less lines and short answer

回答1:

You can't get all days just by declaring the month, you need to add the year as well because of leap years:

DECLARE @date DATE = getdate()

;WITH N(N)AS 
(SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1))M(N)),
tally(N)AS(SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a)
SELECT top(day(EOMONTH(@date)))
  N day,
  dateadd(d,N-1, @date) date
FROM tally

Another different solution(by t@clausen):

DECLARE @month AS INT = 5
DECLARE @Year AS INT = 2016

;WITH N(N)AS 
(SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1))M(N)),
tally(N)AS(SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a)
SELECT N day,datefromparts(@year,@month,N) date FROM tally
WHERE N <= day(EOMONTH(datefromparts(@year,@month,1)))

回答2:

Same approach as t-clausen, but a more compact:

Declare @year int = 2017, @month int = 11;
WITH numbers
as
(
    Select 1 as value
    UNion ALL
    Select value + 1 from numbers
    where value + 1 <= Day(EOMONTH(datefromparts(@year,@month,1)))
)
SELECT datefromparts(@year,@month,numbers.value) Datum FROM numbers

回答3:

If you have a date/time column, then use the month() function:

select t.*
from t
where month(datecol) = 5;

回答4:

You can get all the dates of a month using the following query.

declare @month int, @year int
set @month = 2
set @year = 2008

SELECT
CAST(CAST(@year AS VARCHAR) + '-' + CAST(@Month AS VARCHAR) + '-01' AS DATETIME) + Number
FROM master..spt_values
WHERE type = 'P'
AND
(CAST(CAST(@year AS VARCHAR) + '-' + CAST(@Month AS VARCHAR) + '-01' AS DATETIME) + Number )
<
DATEADD(mm,1,CAST(CAST(@year AS VARCHAR) + '-' + CAST(@Month AS VARCHAR) + '-01' AS DATETIME) )

Hope this will help you.

回答5:

Another CTE based version

DECLARE @Month INT = 2, @Year INT = 2019

;WITH MonthDays_CTE(DayNum)
AS
(
SELECT DATEFROMPARTS(@Year, @Month, 1) AS DayNum
UNION ALL
SELECT DATEADD(DAY, 1, DayNum)
  FROM MonthDays_CTE
  WHERE DayNum < EOMONTH(DATEFROMPARTS(@Year, @Month, 1))
)
SELECT *
  FROM MonthDays_CTE

回答6:

DECLARE @MonthNo TINYINT = 03 -- set your month
    ,@WOYEAR SMALLINT = 2018; -- set your year

IF OBJECT_ID('TEMPDB..#TMP') IS NOT NULL
    DROP TABLE #TMP

DECLARE @START_DATE DATETIME
    ,@END_DATE DATETIME
    ,@CUR_DATE DATETIME

SET @START_DATE = DATEADD(month, @MonthNo - 1, DATEADD(year, @WOYEAR - 1900, 0))
SET @END_DATE = DATEADD(day, - 1, DATEADD(month, @MonthNo, DATEADD(year, @WOYEAR - 1900, 0)))
SET @CUR_DATE = @START_DATE

CREATE TABLE #TMP (
    WEEKDAY VARCHAR(10)
    ,DATE INT
    ,MONTH VARCHAR(10)
    ,YEAR INT
    ,dates VARCHAR(25)
    )

WHILE @CUR_DATE <= @END_DATE
BEGIN
    INSERT INTO #TMP
    SELECT DATENAME(DW, @CUR_DATE)
        ,DATEPART(DAY, @CUR_DATE)
        ,DATENAME(MONTH, @CUR_DATE)
        ,DATEPART(YEAR, @CUR_DATE)
        ,REPLACE(CONVERT(VARCHAR(9), @CUR_DATE, 6), ' ', '-')

    SET @CUR_DATE = DATEADD(DD, 1, @CUR_DATE)
END

SELECT *
FROM #TMP

回答7:

DECLARE @Today DATE= GETDATE() ,
@StartOfMonth DATE ,
@EndOfMonth DATE;

DECLARE @DateList TABLE ( DateLabel VARCHAR(10) );
SET @EndOfMonth = EOMONTH(GETDATE());
SET @StartOfMonth = DATEFROMPARTS(YEAR(@Today), MONTH(@Today), 1);

WHILE @StartOfMonth <= @EndOfMonth
BEGIN
   INSERT  INTO @DateList
   VALUES  ( @StartOfMonth );
   SET @StartOfMonth = DATEADD(DAY, 1, @StartOfMonth);
END;

SELECT  DateLabel
FROM    @DateList; 

回答8:

WHERE Dates LIKE '2018-12%'

In a datetime or timestamp it goes Year-Month, so this would pull everything that matches 2018 in December. You can mod this to use your variables as well.

@month = 12;
@year = 2018;
@searchQuery = @year + @month + '%';

WHERE Dates LIKE @searchQuery

回答9:

Little modification. Query given by @t-clausen.dk will give you correct result only if you ran it on first of the month. With little change this works awesome.

DECLARE @date DATE = getdate()

;WITH N(N)AS 
(SELECT 1 FROM(VALUES(1),(1),(1),(1),(1),(1))M(N)),
tally(N)AS(SELECT ROW_NUMBER()OVER(ORDER BY N.N)FROM N,N a)


SELECT top(day(EOMONTH(@date)))
 N day,
 DATEFROMPARTS(year(@date),month(@date), n) date
FROM tally

BTW nice trick t-clausen.dk. I couldn't think of more easy way