First Day of financial year is April 1st.
T-SQL Query to return April 1st for the getdate()
Financial Year: April 1st to March 31st
Try this:
select DATEFROMPARTS(Yr, 4, 1) [start], DATEFROMPARTS(Yr + 1, 3, 31) [end] from
(select case when DATEPART(month, getdate()) < 4 then DATEPART(year, getdate()) - 1 else DATEPART(year, getdate()) end Yr) a
declare #today date = '2018-06-21'
select fin_year = dateadd(month, 3,
dateadd(year,
datepart(year,
dateadd(month, -3, #today)) - 1900, 0))
the expression datepart(year, dateadd(month, -3, #today)) is to get the current financial year. Since your financial year is Apr 1, for Jan 1 to Mar 31, subtracting 3 months from it, will give you the correct year (fiscal year = financial year).
After that it is just to form the date Apr 1 with that year
DECLARE #DateToUse DATETIME = GETDATE(),
#FinancialYearStart DATETIME
DECLARE #DayPart INT = DATEPART(DAY, #DateToUse),
#MonthPart INT = DATEPART(MONTH, #DateToUse),
#YearPart INT = DATEPART(YEAR, #DateToUse),
#StartMonth INT = 4, -- April
#StartDay INT = 1 -- 1st
SELECT DATETIMEFROMPARTS((
#YearPart - CASE
WHEN #MonthPart > #StartMonth
OR (
#MonthPart = #StartMonth
AND #DayPart >= #StartDay
)
THEN 0
ELSE 1
END
), #StartMonth, #StartDay, 0, 0, 0, 0)
I just knocked this up and it produces 2020-04-01 00:00:00.000 right now. I threw some other dates at it and it seemed to work nicely. Obviously the month/day can easily be changed by changing the variables at the top.
It's not the shortest block of SQL, but it's easy to use for different dates. A lot of other answers I've seen across different questions have returned different formats like a string or just the year part. Whereas this datetime will allow it to be used for datetime comparisons.
Number of days left in a given month
How do I find the number of days left in the current month?
Example if current month is November and todays date is 16/11/2016
The Numbers of days in month – Elapse days = ? I want to do it dynamically
In my example 30 – 16 = 14
declare #date date
set #date='16 Nov 2016'
select datediff(day, #date, dateadd(month, 1, #date)) - 16 AS DaysLeft
Since this is sql server 2008 you can't use EOMonth (that was introduced in 2012 version).
You have to do some date adds and date diffs:
SELECT DATEDIFF(DAY,
GETDATE(),
DATEADD(MONTH,
1,
DATEADD(DAY, 1 - DAY(GETDATE()), GETDATE())
)
) - 1
explanations:
DATEADD(DAY, 1 - DAY(GETDATE()), GETDATE()) gets the first day of the current month, the wrapping DATEADD adds one month, and the wrapping DATEDIFF returns the number of days between the current date and the first date of the next month. This is why you need to subtruct 1 to get the correct number of days.
--For SQL 2012 And Above Version Use Below Query to Get Count Of Days Left In A Month
DECLARE #date DATE
SET #date=GETDATE()
SELECT DATEDIFF(DAY, #date,EOMONTH(#date))
-- And for Sql Server 2008 Use Below Query to Get Count of Days Left for the Month
DECLARE #date Date
SET #date=GETDATE()
SELECT DATEDIFF(DAY, #date, DATEADD(MONTH, 1, #date)) - DATENAME(DAY,GETDATE())
AS DaysLeft
Simply use the Datepart function:
declare #date date
set #date='16 Nov 2016'
select datediff(day, #date, dateadd(month, 1, #date)) - Datepart(DAY,#date)
Change your date to getdate()
declare #date date
set #date=GETDATE()
select datediff(day, #date, dateadd(month, 1, #date)) - DATENAME(DAY,GETDATE())
AS DaysLeft
DECLARE #date DATE SET #date='16 Nov 2016' SELECT DATEDIFF(DAY, #date,EOMONTH(#date))
Use EOMONTH and DATEDIFF functions
declare #date date
set #date='16 Nov 2016'
select datediff(day,#date,eomonth(#date)) as days_left
Use below solution For Below versions of sql server 2012
DECLARE #DATE DATE
SET #DATE='16 NOV 2016'
SELECT DATEDIFF(DAY,#DATE,DATEADD( MM,DATEDIFF(MM,0,#DATE)+1,0))-1 AS DAYS_LEFT
FUNCTION example
CREATE Or ALTER FUNCTION ReturnDaysLeftInMonth(#Date Date)
RETURNS Int
AS
BEGIN
RETURN DATEDIFF(DAY, #Date, EOMONTH(#Date)) + 1
END
Or use
Declare #Date Date
Set #Date=GETDATE()
DATEDIFF(DAY, #Date, EOMONTH(#Date)) + 1
I have searched far and wide, but I can't seem find a way to convert julian to yyyy-mm-dd.
Here is the format of my julian:
The Julian format consists of the year, the first two digits, and the day within the year, the last three digits.
For example, 95076 is March 17, 1995. The 95 indicates the year and the
076 indicates it is the 76th day of the year.
15260
I have tried this but it isn't working:
dateadd(d,(convert(int,LAST_CHANGED_DATE) % 1000)-1, convert(date,(convert(varchar,convert(int,LAST_CHANGED_DATE) /1000 + 1900) + '/1/1'))) as GrgDate
You can select each part of the date using datepart()
SELECT DATEPART(yy, 95076), DATEPART(dy, 95076)
+++EDIT: I misunderstood something. Here's my correction: +++++
SELECT DATEADD(day, CAST(RIGHT('95076',3) AS int) – 1, CONVERT(datetime,LEFT('95076',2) + '0101', 112))
Edit: leaving this answer for Oracle and MySQL users
This will not work in T-SQL.
Use this:
MAKEDATE(1900 + d / 1000, d % 1000)
For example:
SELECT MAKEDATE(1900 + 95076 / 1000, 95076 % 1000)
This returns March, 17 1995 00:00:00.
SQLFiddle
I concatenated 20 to my JD and then ran
DATEADD(YEAR, LAST_CHANGE_DATE / 1000 - 1900, LAST_CHANGE_DATE % 1000 - 1)
this got me the result. Thank you!!!
FOR SQL Users
DECLARE #jdate VARCHAR(10)
SET #jdate = 117338
SELECT dateadd(dd, (#jdate - ((#jdate/1000) * 1000)) - 1, dateadd(yy, #jdate/1000, 0))
This will definitely work in all case.
DECLARE #date int
SET #date = 21319
SELECT DATEADD(dd, RIGHT(#date,LEN(#date)-3)-1, DATEADD(yy,LEFT(#date,1)*100 +RIGHT(LEFT(#date,3),2),'1 Jan 1900'))
You need to specify 1 or 0 for century in first character. For example, SET #date = 21319 should be prefixed with 1 or 0. Below is an example that will work with all y2k use cases.
DECLARE #jdate INT
SET #jdate = 119150
SELECT DATEADD(dd, (#jdate - ((#jdate/1000) * 1000)) - 1, DATEADD(yy, #jdate/1000, 0))
Declare #Julian varchar(7)
Declare #date date
Set #Julian = 2020277
Set #Date = Dateadd(day,+ Cast(right(#Julian,3) as int)-1, Cast(left(#Julian,4) + '0101' as Date))
Select #Date
Standard SQL is simple (you can do similar for 2 digit year YYMMDD)
Declare #julDate int = 2020275
Select
DateAdd
(
day
, Right(#julDate,3)-1
, Cast((Left(#julDate,4)+'-01-01') as smalldatetime)
)
Try this out:
DECLARE #jdate int
SET #jdate = 21243
select dateadd(dd, (#jdate - ((#jdate/1000) * 1000)) - 1, dateadd(yy, #jdate/1000+100, 0))
Got 2 parameters #yr and #period, #period is just the month number so July would equal 7 for example.
In my stored procedure table I've got a column called Date which is just a standard datetime field. I need a where clause to work out all dates greater than the current period minus 1 year so if #period = 7 and #yr = 2012 I want the where clause to return all dates greater than '01-07-2011' (UK date format) how can I achieve this with just the 2 numbers from #period and #yr.
WHERE <br>
Date >= '01-07-2011'
You could
Date >= dateadd(month, #period-1, dateadd(year, #yr-1900, 0))
where year(date)>year(getdate()-1) and month(date)>#period
If you want the expression sargable, convert it to datetime:
declare #year int = 2012
declare #month int = 7
select
...
where [Date] >= convert(datetime, convert(varchar(4), #year)
+ right('0' + convert (varchar(2), #month), 2)
+ '01')
After seeing Alex K.'s answer, you might even do this:
dateadd(month, #month - 1 + (#year-1900) * 12, 0)
For the best performance you should do something like this:
declare #yr int = 2012
declare #period int = 7
select ...
from ....
WHERE date >= dateadd(month, (#yr - 1901) * 12 + #period - 1, 0)
We can do it in may ways
try it
DECLARE #a VARCHAR(20),
#b VARCHAR(10),
#c varchar(4)
SET #b='may' /*pass your stored proc value */
SET #c='2011'
SET #a='01'+#b+#c
SET DATEFORMAT YDM
SELECT CAST(#a AS DATE)
FOR uk formate
SELECT CONVERT(char,CAST(#a AS DATE),103)
Just t make sure you compare against an entire date, one solution I'd offer is:
Select *
from TheTable
where date> DateAdd(Year,-1, convert(datetime, '01/'+convert(varchar,#period)+'/' + convert(varchar,#yr)))
To account for regional format differences in SQL Server 2012:
Select *
from TheTable
where date> DateAdd(Year,-1, DateFromParts(#year,#period,1))
For pre-2012:
Select *
from TheTable
Where Date > DateAdd(day, 0, DateAdd(month, #period-1, DateAdd(year, (#yr-1900)-1,0)))
The #yr-1900 is maintained to illustrate the computation of the base date offset from 1900, then subtracting 1 for the one-year-off date computation
I need to determine the number of days in a month for a given date in SQL Server.
Is there a built-in function? If not, what should I use as the user-defined function?
In SQL Server 2012 you can use EOMONTH (Transact-SQL) to get the last day of the month and then you can use DAY (Transact-SQL) to get the number of days in the month.
DECLARE #ADate DATETIME
SET #ADate = GETDATE()
SELECT DAY(EOMONTH(#ADate)) AS DaysInMonth
You can use the following with the first day of the specified month:
datediff(day, #date, dateadd(month, 1, #date))
To make it work for every date:
datediff(day, dateadd(day, 1-day(#date), #date),
dateadd(month, 1, dateadd(day, 1-day(#date), #date)))
Most elegant solution: works for any #DATE
DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,#DATE),0)))
Throw it in a function or just use it inline. This answers the original question without all the extra junk in the other answers.
examples for dates from other answers:
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'1/31/2009'),0))) Returns 31
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2404-feb-15'),0))) Returns 29
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2011-12-22'),0))) Returns 31
--Last Day of Previous Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0)))
--Last Day of Current Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+1,0)))
--Last Day of Next Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+2,0)))
Personally though, I would make a UDF for it if there is not a built in function...
I would suggest:
SELECT DAY(EOMONTH(GETDATE()))
This code gets you the number of days in current month:
SELECT datediff(dd,getdate(),dateadd(mm,1,getdate())) as datas
Change getdate() to the date you need to count days for.
--- sql server below 2012---
select day( dateadd(day,-1,dateadd(month, 1, convert(date,'2019-03-01'))))
-- this for sql server 2012--
select day(EOMONTH(getdate()))
Solution 1: Find the number of days in whatever month we're currently in
DECLARE #dt datetime
SET #dt = getdate()
SELECT #dt AS [DateTime],
DAY(DATEADD(mm, DATEDIFF(mm, -1, #dt), -1)) AS [Days in Month]
Solution 2: Find the number of days in a given month-year combo
DECLARE #y int, #m int
SET #y = 2012
SET #m = 2
SELECT #y AS [Year],
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m - 1, 0)),
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m, 0))
) AS [Days in Month]
You do need to add a function, but it's a simple one. I use this:
CREATE FUNCTION [dbo].[ufn_GetDaysInMonth] ( #pDate DATETIME )
RETURNS INT
AS
BEGIN
SET #pDate = CONVERT(VARCHAR(10), #pDate, 101)
SET #pDate = #pDate - DAY(#pDate) + 1
RETURN DATEDIFF(DD, #pDate, DATEADD(MM, 1, #pDate))
END
GO
SELECT Datediff(day,
(Convert(DateTime,Convert(varchar(2),Month(getdate()))+'/01/'+Convert(varchar(4),Year(getdate())))),
(Convert(DateTime,Convert(varchar(2),Month(getdate())+1)+'/01/'+Convert(varchar(4),Year(getdate()))))) as [No.of Days in a Month]
select datediff(day,
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3 - 1, 0)),
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3, 0))
)
Nice Simple and does not require creating any functions Work Fine
You need to create a function, but it is for your own convenience. It works perfect and I never encountered any faulty computations using this function.
CREATE FUNCTION [dbo].[get_days](#date datetime)
RETURNS int
AS
BEGIN
SET #date = DATEADD(MONTH, 1, #date)
DECLARE #result int = (select DAY(DATEADD(DAY, -DAY(#date), #date)))
RETURN #result
END
How it works: subtracting the date's day number from the date itself gives you the last day of previous month. So, you need to add one month to the given date, subtract the day number and get the day component of the result.
select add_months(trunc(sysdate,'MM'),1) - trunc(sysdate,'MM') from dual;
I upvoted Mehrdad, but this works as well. :)
CREATE function dbo.IsLeapYear
(
#TestYear int
)
RETURNS bit
AS
BEGIN
declare #Result bit
set #Result =
cast(
case when ((#TestYear % 4 = 0) and (#testYear % 100 != 0)) or (#TestYear % 400 = 0)
then 1
else 0
end
as bit )
return #Result
END
GO
CREATE FUNCTION dbo.GetDaysInMonth
(
#TestDT datetime
)
RETURNS INT
AS
BEGIN
DECLARE #Result int
DECLARE #MonthNo int
Set #MonthNo = datepart(m,#TestDT)
Set #Result =
case #MonthNo
when 1 then 31
when 2 then
case
when dbo.IsLeapYear(datepart(yyyy,#TestDT)) = 0
then 28
else 29
end
when 3 then 31
when 4 then 30
when 5 then 31
when 6 then 30
when 7 then 31
when 8 then 31
when 9 then 30
when 10 then 31
when 11 then 30
when 12 then 31
end
RETURN #Result
END
GO
To Test
declare #testDT datetime;
set #testDT = '2404-feb-15';
select dbo.GetDaysInMonth(#testDT)
here's another one...
Select Day(DateAdd(day, -Day(DateAdd(month, 1, getdate())),
DateAdd(month, 1, getdate())))
I know this question is old but I thought I would share what I'm using.
DECLARE #date date = '2011-12-22'
/* FindFirstDayOfMonth - Find the first date of any month */
-- Replace the day part with -01
DECLARE #firstDayOfMonth date = CAST( CAST(YEAR(#date) AS varchar(4)) + '-' +
CAST(MONTH(#date) AS varchar(2)) + '-01' AS date)
SELECT #firstDayOfMonth
and
DECLARE #date date = '2011-12-22'
/* FindLastDayOfMonth - Find what is the last day of a month - Leap year is handled by DATEADD */
-- Get the first day of next month and remove a day from it using DATEADD
DECLARE #lastDayOfMonth date = CAST( DATEADD(dd, -1, DATEADD(mm, 1, FindFirstDayOfMonth(#date))) AS date)
SELECT #lastDayOfMonth
Those could be combine to create a single function to retrieve the number of days in a month if needed.
SELECT DAY(SUBDATE(ADDDATE(CONCAT(YEAR(NOW()), '-', MONTH(NOW()), '-1'), INTERVAL 1 MONTH), INTERVAL 1 DAY))
Nice 'n' Simple and does not require creating any functions
Mehrdad Afshari reply is most accurate one, apart from usual this answer is based on formal mathematical approach given by Curtis McEnroe in his blog https://cmcenroe.me/2014/12/05/days-in-month-formula.html
DECLARE #date DATE= '2015-02-01'
DECLARE #monthNumber TINYINT
DECLARE #dayCount TINYINT
SET #monthNumber = DATEPART(MONTH,#date )
SET #dayCount = 28 + (#monthNumber + floor(#monthNumber/8)) % 2 + 2 % #monthNumber + 2 * floor(1/#monthNumber)
SELECT #dayCount + CASE WHEN #dayCount = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END -- leap year adjustment
To get the no. of days in a month we can directly use Day() available in SQL.
Follow the link posted at the end of my answer for SQL Server 2005 / 2008.
The following example and the result are from SQL 2012
alter function dbo.[daysinm]
(
#dates nvarchar(12)
)
returns int
as
begin
Declare #dates2 nvarchar(12)
Declare #days int
begin
select #dates2 = (select DAY(EOMONTH(convert(datetime,#dates,103))))
set #days = convert(int,#dates2)
end
return #days
end
--select dbo.daysinm('08/12/2016')
Result in SQL Server SSMS
(no column name)
1 31
Process:
When EOMONTH is used, whichever the date format we use it is converted into DateTime format of SQL-server. Then the date output of EOMONTH() will be 2016-12-31 having 2016 as Year, 12 as Month and 31 as Days.
This output when passed into Day() it gives you the total days count in the month.
If we want to get the instant result for checking we can directly run the below code,
select DAY(EOMONTH(convert(datetime,'08/12/2016',103)))
or
select DAY(EOMONTH(convert(datetime,getdate(),103)))
for reference to work in SQL Server 2005/2008/2012, please follow the following external link ...
Find No. of Days in a Month in SQL
DECLARE #date DATETIME = GETDATE(); --or '12/1/2018' (month/day/year)
SELECT DAY(EOMONTH ( #date )) AS 'This Month';
SELECT DAY(EOMONTH ( #date, 1 )) AS 'Next Month';
result:
This Month
31
Next Month
30
DECLARE #m int
SET #m = 2
SELECT
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, +#m -1, 0)),
DATEADD(DAY, 0, DATEADD(m,+ #m, 0))
) AS [Days in Month]
RETURN day(dateadd(month, 12 * #year + #month - 22800, -1))
select day(dateadd(month, 12 * year(date) + month(date) - 22800, -1))
A cleaner way of implementing this is using the datefromparts function to construct the first day of the month, and calculate the days from there.
CREATE FUNCTION [dbo].[fn_DaysInMonth]
(
#year INT,
#month INT
)
RETURNS INT
AS
BEGIN
IF #month < 1 OR #month > 12 RETURN NULL;
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, #month, 1);
DECLARE #lastDay DATE = dateadd(month, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
Similarily, you can calculate the days in a year:
CREATE FUNCTION [dbo].[fn_DaysInYear]
(
#year INT
)
RETURNS INT
AS
BEGIN
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, 1, 1);
DECLARE #lastDay DATE = dateadd(year, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
use SQL Server EOMONTH Function nested with day to get last day of month
select Day(EOMONTH('2020-02-1')) -- Leap Year returns 29
select Day(EOMONTH('2021-02-1')) -- returns 28
select Day(EOMONTH('2021-03-1')) -- returns 31
For any date
select DateDiff(Day,#date,DateAdd(month,1,#date))
select first_day=dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),
last_day=dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())),
no_of_days = 1+datediff(dd,dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())))
replace any date with getdate to get the no of months in that particular date
DECLARE #Month INT=2,
#Year INT=1989
DECLARE #date DateTime=null
SET #date=CAST(CAST(#Year AS nvarchar) + '-' + CAST(#Month AS nvarchar) + '-' + '1' AS DATETIME);
DECLARE #noofDays TINYINT
DECLARE #CountForDate TINYINT
SET #noofDays = DATEPART(MONTH,#date )
SET #CountForDate = 28 + (#noofDays + floor(#noofDays/8)) % 2 + 2 % #noofDays + 2 * floor(1/#noofDays)
SET #noofDays= #CountForDate + CASE WHEN #CountForDate = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END
PRINT #noofDays
DECLARE #date nvarchar(20)
SET #date ='2012-02-09 00:00:00'
SELECT DATEDIFF(day,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime),dateadd(month,1,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime)))
simple query in SQLServer2012 :
select day(('20-05-1951 22:00:00'))
i tested for many dates and it return always a correct result