Number of days left in current month - sql

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

Related

How to get number of days in a month in SQL Server

Is there a way to get the number of days in a month in SQL Server, if we input the month number or month name, or even a date?

You can use:
select day(eomonth ('2018-02-01')) as NoOfDays
and the result will be:
NoOfDays
-----------
28

If you have a date, then simply do:
select day(eomonth(date))
If you have a month number, then:
select day(eomonth(datefromparts(2020, month_number, 1)))

If you have a date and are on 2012 or later :
SELECT day(eomonth(yourdate))
Month name / number is automatically prone to an error when dealing with February - do you consider it 28 or 29, which year are you referring to when making that calculation etc.

In case you are using sql-server 2008 or earlier:
Date as input
DECLARE #date DATETIME = getdate()
SELECT day(dateadd(m, datediff(m, -1, #date), -1))
Month and year as input
DECLARE #year INT = 2018
DECLARE #month INT = 2
SELECT day(dateadd(m, #month + datediff(m, 0, cast(#year as char(4))), -1))

try using
SELECT day(eomonth(yourdate))

for pre 2012 where eomonth() is not available
if you have a date
select day(dateadd(month, datediff(month, 0, #date) + 1, -1))
if you have the year & month
declare #year int = 2018,
#month int = 8
select dateadd(month, #month, dateadd(year, #year - 1900, -1))

tsql: How to retrieve the last date of each month between given date range

I have two date for example 08/08/2013 and 11/11/2013 and I need last date of each month starting from August to November in a table so that i can iterate over the table to pick those dates individually.
I know how to pick last date for any month but i am stucked with a date range.
kindly help, it will be highly appreciated.
Note : I am using Sql 2008 and date rang could be 1 month , 2 month or 6 month or a year or max too..

You can use CTE for getting all last days of the month within the defined range
Declare #Start datetime
Declare #End datetime
Select #Start = '20130808'
Select #End = '20131111'
;With CTE as
(
Select #Start as Date,Case When DatePart(mm,#Start)<>DatePart(mm,#Start+1) then 1 else 0 end as [Last]
UNION ALL
Select Date+1,Case When DatePart(mm,Date+1)<>DatePart(mm,Date+2) then 1 else 0 end from CTE
Where Date<#End
)
Select * from CTE
where [Last]=1 OPTION ( MAXRECURSION 0 )

DECLARE #tmpTable table (LastDates DATE);
DECLARE #startDate DATE = '01/01/2012'; --1 Jan 2012
DECLARE #endDate DATE = '05/31/2012'; --31 May 2012
DECLARE #tmpEndDate DATE;
SET #startDate = DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#startDate)+1,1));
SET #tmpEndDate = DATEADD(DAY, 1, #endDate);
WHILE (#startDate <= #tmpEndDate)
BEGIN
INSERT INTO #tmpTable (LastDates) values (DATEADD(DAY, -1, #startDate));
SET #startDate = DATEADD(MONTH, 1, #startDate);
END
SELECT [LastDates] FROM #tmpTable;
Output:
Example: 1
#startDate DATE = '01/01/2012'; --1 Jan 2012
#endDate DATE = '05/31/2012'; --31 May 2012
LastDates
----------
2012-01-31
2012-02-29
2012-03-31
2012-04-30
2012-05-31
Example: 2
#startDate DATE = '11/01/2011'; --1 Nov 2011
#endDate DATE = '03/13/2012'; --13 Mar 2012
LastDates
----------
2011-11-30
2011-12-31
2012-01-31
2012-02-29

I've created a table variable, filled it with all days between #startDate and #endDate and searched for max date in the month.
declare #tmpTable table (dates date)
declare #startDate date = '08/08/2013'
declare #endDate date = '11/11/2013'
while #startDate <= #endDate
begin
insert into #tmpTable (dates) values (#startDate)
set #startDate = DATEADD(DAY, 1, #startDate)
end
select max(dates) as [Last day] from #tmpTable as o
group by datepart(YEAR, dates), datepart(MONTH, dates)
Results:
Last day
2013-08-31
2013-09-30
2013-10-31
2013-11-11
To also get last day of November this can be used before loop:
set #endDate = DATEADD(day, -1, DATEADD(month, DATEDIFF(month, 0, #endDate) + 1, 0))

Following script demonstrates the script to find last day of previous, current and next month.
----Last Day of Previous Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0))
LastDay_PreviousMonth
----Last Day of Current Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+1,0))
LastDay_CurrentMonth
----Last Day of Next Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+2,0))
LastDay_NextMonth
If you want to find last day of month of any day specified use following script.
--Last Day of Any Month and Year
DECLARE #dtDate DATETIME
SET #dtDate = '8/18/2007'
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#dtDate)+1,0))
LastDay_AnyMonth
ResultSet:
LastDay_AnyMonth
Source - SQL Server Central.

You can use a recursive CTE to do this, note the MAXRECURSION OPTION prevents an infinite loop:
DECLARE #StartDate DATE = '2013-08-08'
DECLARE #EndDate DATE = '2013-11-11'
;WITH dateCTE
AS
(
SELECT CAST(DATEADD(M, 1,DATEADD(d, DAY(#StartDate) * -1, #StartDate)) AS DATE) EndOFMonth
UNION ALL
SELECT CAST(DATEADD(M, 2,DATEADD(d, DAY(EndOFMonth) * -1, EndOFMonth)) AS DATE)
FROM dateCTE
WHERE EndOFMonth < DATEADD(d, DAY(#EndDate) * -1, #EndDate)
)
SELECT *
FROM dateCTE
OPTION (MAXRECURSION 30);
This returns
EndOFMonth
----------
2013-08-31
2013-09-30
2013-10-31

try this
the last row(where) is optional for date filtering
declare #table table
(
thisdate date
)
insert into #table values ('12/01/2013'),('05/06/2013'),('04/29/2013'),('02/20/2013')
select *,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,thisdate)+1,0))
LastDay from #table
where thisdate between 'givendate' and 'givendate'
The Example Below is for all dates
thisdate lastday
2013-12-01 2013-12-31 23:59:59.000
2013-05-06 2013-05-31 23:59:59.000
2013-04-29 2013-04-30 23:59:59.000
2013-02-20 2013-02-28 23:59:59.000

The following CTE gives you the last day of every month from February 1900 until the middle of the 26th century (on my machine):
;with LastDaysOfMonths as (
select DATEADD(month,
ROW_NUMBER() OVER (ORDER BY so.object_id),
'19000131') as Dt
from sys.objects so,sys.objects so1
)
select * from LastDaysOfMonths
It should be easy enough to use it as part of a larger query or to filter it down to just the dates you want. You can adjust the range of years as needed by changing the constant 19000131. The only important thing to do is make sure that you use a month that has 31 days in it and always have the constant be for day 31.

No need to use a common table expression or anything like that - this simple query will do it:
SELECT DATEADD(d, -1, DATEADD(mm, DATEDIFF(m, 0, DATEADD(m, number, '2013-08-08')) + 1, 0)) AS EndOfMonth
FROM master.dbo.spt_values
WHERE 'P' = type
AND DATEADD(m, number, '2013-08-08') < '2013-11-11';

Although the question is about the last day which #bummi has already answered.
But here is the solution for the first date which might be helpful for someone.
Get the first dates of all the months in-between the #FromDate and #ToDate.
DECLARE #FromDate DATETIME = '2019-08-13'
DECLARE #ToDate DATETIME = '2019-11-25'
;WITH CTE
AS
(
SELECT DATEADD(DAY, -(DAY(#FromDate) - 1), #FromDate) AS FirstDateOfMonth
UNION ALL
SELECT DATEADD(MONTH, 1, FirstDateOfMonth)
FROM CTE
WHERE FirstDateOfMonth < DATEADD(DAY, -(DAY(#ToDate) - 1), #ToDate)
)
SELECT * FROM CTE
Here is the result
--Result
2019-08-01 00:00:00.000
2019-09-01 00:00:00.000
2019-10-01 00:00:00.000
2019-11-01 00:00:00.000

Date Value from a week number in SQL [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Get dates from a week number in T-SQL
How do I get the date value if I have a week number in SQL Query.
Like if I pass 26, it should give me 06/24/2012. If I pass 27, I should get 07/01/2012
Any help will be appreciated :)
Sots

SELECT DATEADD(week, n, '11/25/2011');
with n being the week number

If this doesn't work, try using WEEK() instead of WEEKOFYEAR().
CURDATE() - INTERVAL WEEKDAY(CURDATE()) DAY + INTERVAL (WEEKNO - WEEKOFYEAR(CURDATE())) WEEK

In SQL Server
DECLARE #StartDate DATE, #WeekVal INT
SET #WeekVal = 26 -- Set the week number
SET #StartDate = DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0) -- Start of current year
;WITH cte AS (
SELECT #StartDate AS DateVal, DATEPART(wk, #StartDate) AS WeekVal, 1 AS RowVal
UNION ALL
SELECT DATEADD(d, 1, DateVal), DATEPART(wk, DATEADD(d, 1, DateVal)), RowVal + 1
FROM cte WHERE RowVal < 365
)
SELECT MIN(DateVal) StartOfWeek
FROM cte
WHERE WeekVal = #WeekVal
OPTION (MAXRECURSION 365);

This gives you the start and end dates of the week. [For SQL Server]
Declare #week integer set #week = 26
Declare #Year Integer Set #Year = year(getdate())
declare #date datetime
-- ------------------------------------
Set #date = DateAdd(day, 0,
DateAdd(month, 0,
DateAdd(Year, #Year-1900, 0)))
set #date = Dateadd(week, #week-1, #date)
select #date startweek, DATEADD (D, -1 * DatePart (DW, #date) + 7, #date) endweek
This was the result from it:
startweek endweek
----------------------- -----------------------
2012-07-01 00:00:00.000 2012-07-07 00:00:00.000
(1 row(s) affected)

Playing with Date Time in SQL Server

I am building an SSRS report.
In the report, the week always runs from Monday - Sunday.
And I want to find out the START and END dates of prior two weeks.
For example,
If current date = Jan 1, 2011
-- current week = Dec 27, 2010 to Jan 2, 2011
-- previous week = Dec 20, 2010 to Dec 26, 2010
I have tried the following, but seems like it fails when when current day = Sunday
DECLARE #DT DATETIME
DECLARE #Offset INT
DECLARE #CM DATETIME
DECLARE #PM DATETIME
DECLARE #PS DATETIME
--SET #DT = GETDATE()
SET #DT = '11/14/2010' -- Monday
SET #Offset = (DATEPART(WEEKDAY, #DT) - 2) * -1
SET #CM = DATEADD(DAY, #Offset, #DT)
SET #PM = DATEADD(DAY, -7, #CM)
SET #PS = DATEADD(DAY, -1, #CM)
SELECT #Offset AS Offset, #DT AS Date, #CM AS Monday, #PM AS [Previous Monday], #PS AS [Previous Sunday], DATEPART(WK, #PM) AS Week
How can I fix it?

Add a [Calendar] table to your DB with all the dates you need precalculated. You can then include fields with the day name, number, holiday etc. and simply look up the values you need.
It's much simpler than playing with DATEADD
A useful article on calendar tables: Why should I consider using an auxiliary calendar table?

Not sure if this is the most elegant solution but you could do a case statement with your #offset like this:
SET #offset = CASE WHEN DATEPART(weekday, #today) >= 2
THEN -(DATEPART(weekday, #today) - 2)
ELSE -(DATEPART(weekday, #today) + 5)
END
I believe it works for all cases.

Get the last day of the month in SQL

I need to get the last day of the month given as a date in SQL. If I have the first day of the month, I can do something like this:
DATEADD(DAY, DATEADD(MONTH,'2009-05-01',1), -1)
But does anyone know how to generalize it so I can find the last day of the month for any given date?

From SQL Server 2012 you can use the EOMONTH function.
Returns the last day of the month that contains the specified date,
with an optional offset.
Syntax
EOMONTH ( start_date [, month_to_add ] )
How ... I can find the last day of the month for any given date?
SELECT EOMONTH(#SomeGivenDate)

Here's my version. No string manipulation or casting required, just one call each to the DATEADD, YEAR and MONTH functions:
DECLARE #test DATETIME
SET #test = GETDATE() -- or any other date
SELECT DATEADD(month, ((YEAR(#test) - 1900) * 12) + MONTH(#test), -1)

You could get the days in the date by using the DAY() function:
dateadd(day, -1, dateadd(month, 1, dateadd(day, 1 - day(date), date)))

Works in SQL server
Declare #GivenDate datetime
SET #GivenDate = GETDATE()
Select DATEADD(MM,DATEDIFF(MM, 0, #GivenDate),0) --First day of the month
Select DATEADD(MM,DATEDIFF(MM, -1, #GivenDate),-1) --Last day of the month

I know this is a old question but here is another solution that works for me
SET #dtDate = "your date"
DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#dtDate)+1,0))
And if some one is looking for different examples here is a link http://blog.sqlauthority.com/2007/08/18/sql-server-find-last-day-of-any-month-current-previous-next/
I hope this helps some one else.
stackoverflow Rocks!!!!

For SQL server 2012 or above use EOMONTH to get the last date of month
SQL query to display end date of current month
DECLARE #currentDate DATE = GETDATE()
SELECT EOMONTH (#currentDate) AS CurrentMonthED
SQL query to display end date of Next month
DECLARE #currentDate DATE = GETDATE()
SELECT EOMONTH (#currentDate, 1 ) AS NextMonthED

Based on the statements:
SELECT DATEADD(MONTH, 1, #x) -- Add a month to the supplied date #x
and
SELECT DATEADD(DAY, 0 - DAY(#x), #x) -- Get last day of month previous to the supplied date #x
how about adding a month to date #x and then retrieving the last day of the month previous to that (i.e. The last day of the month of the supplied date)
DECLARE #x DATE = '20-Feb-2012'
SELECT DAY(DATEADD(DAY, 0 - DAY(DATEADD(MONTH, 1, #x)), DATEADD(MONTH, 1, #x)))
Note: This was test using SQL Server 2008 R2

Just extend your formula out a little bit:
dateadd(day, -1,
dateadd(month, 1,
cast(month('5/15/2009') as varchar(2)) +
'/1/' +
cast(year('5/15/2009') as varchar(4)))

This works for me, using Microsoft SQL Server 2005:
DATEADD(d,-1,DATEADD(mm, DATEDIFF(m,0,'2009-05-01')+1,0))

WinSQL to get last day of last month (i.e today is 2017-02-09, returns 2017-01-31:
Select dateadd(day,-day(today()),today())

Try to run the following query, it will give you everything you want :)
Declare #a date =dateadd(mm, Datediff(mm,0,getdate()),0)
Print('First day of Current Month:')
Print(#a)
Print('')
set #a = dateadd(mm, Datediff(mm,0,getdate())+1,-1)
Print('Last day of Current Month:')
Print(#a)
Print('')
Print('First day of Last Month:')
set #a = dateadd(mm, Datediff(mm,0,getdate())-1,0)
Print(#a)
Print('')
Print('Last day of Last Month:')
set #a = dateadd(mm, Datediff(mm,0,getdate()),-1)
Print(#a)
Print('')
Print('First day of Current Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate()),0)
Print(#a)
Print('')
Print('Last day of Current Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate())+1,-1)
Print(#a)
Print('')
Print('First day of Last Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate())-1,0)
Print(#a)
Print('')
Print('Last day of Last Week:')
set #a = dateadd(ww, Datediff(ww,0,getdate()),-1)
Print(#a)

WinSQL: I wanted to return all records for last month:
where DATE01 between dateadd(month,-1,dateadd(day,1,dateadd(day,-day(today()),today()))) and dateadd(day,-day(today()),today())
This does the same thing:
where month(DATE01) = month(dateadd(month,-1,today())) and year(DATE01) = year(dateadd(month,-1,today()))

This query can also be used.
DECLARE #SelectedDate DATE = GETDATE()
SELECT DATEADD(DAY, - DAY(#SelectedDate), DATEADD(MONTH, 1 , #SelectedDate)) EndOfMonth

--## Useful Date Functions
SELECT
GETDATE() AS [DateTime],
CAST(GETDATE() AS DATE) AS [Date],
DAY(GETDATE()) AS [Day of Month],
FORMAT(GETDATE(),'MMMM') AS [Month Name],
FORMAT(GETDATE(),'MMM') AS [Month Short Name],
FORMAT(GETDATE(),'MM') AS [Month No],
YEAR(GETDATE()) AS [Year],
CAST(DATEADD(DD,-(DAY(GETDATE())-1),GETDATE()) AS DATE) AS [Month Start Date],
EOMONTH(GETDATE()) AS [Month End Date],
CAST(DATEADD(M,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Previous Month Start Date],
CAST(DATEADD(S,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Previous Month End Date],
CAST(DATEADD(M,+1,DATEADD(MM, DATEDIFF(M,0,GETDATE()),0)) AS DATE) AS [Next Month Start Date],
CAST(DATEADD(D,-1,DATEADD(MM, DATEDIFF(M,0,GETDATE())+2,0)) AS DATE) AS [Next Month End Date],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE()),0) AS DATE) AS [First Day of Current Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+1,-1) AS DATE) AS [Last Day of Current Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())-1,0) AS DATE) AS [First Day of Last Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE()),-1) AS DATE) AS [Last Day of Last Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+1,0) AS DATE) AS [First Day of Next Week],
CAST(DATEADD(WW, DATEDIFF(WW,0,GETDATE())+2,-1) AS DATE) AS [Last Day of Next Week]

My 2 cents:
select DATEADD(DAY,-1,DATEADD(MONTH,1,DATEADD(day,(0-(DATEPART(dd,'2008-02-12')-1)),'2008-02-12')))
Raj

using sql server 2005, this works for me:
select dateadd(dd,-1,dateadd(mm,datediff(mm,0,YOUR_DATE)+1,0))
Basically, you get the number of months from the beginning of (SQL Server) time for YOUR_DATE. Then add one to it to get the sequence number of the next month. Then you add this number of months to 0 to get a date that is the first day of the next month. From this you then subtract a day to get to the last day of YOUR_DATE.

Take some base date which is the 31st of some month e.g. '20011231'. Then use the
following procedure (I have given 3 identical examples below, only the #dt value differs).
declare #dt datetime;
set #dt = '20140312'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');
set #dt = '20140208'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');
set #dt = '20140405'
SELECT DATEADD(month, DATEDIFF(month, '20011231', #dt), '20011231');

Using SQL Server, here is another way to find last day of month :
SELECT DATEADD(MONTH,1,GETDATE())- day(DATEADD(MONTH,1,GETDATE()))

I wrote following function, it works.
It returns datetime data type. Zero hour, minute, second, miliseconds.
CREATE Function [dbo].[fn_GetLastDate]
(
#date datetime
)
returns datetime
as
begin
declare #result datetime
select #result = CHOOSE(month(#date),
DATEADD(DAY, 31 -day(#date), #date),
IIF(YEAR(#date) % 4 = 0, DATEADD(DAY, 29 -day(#date), #date), DATEADD(DAY, 28 -day(#date), #date)),
DATEADD(DAY, 31 -day(#date), #date) ,
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date),
DATEADD(DAY, 30 -day(#date), #date),
DATEADD(DAY, 31 -day(#date), #date))
return convert(date, #result)
end
It's very easy to use.
2 example:
select [dbo].[fn_GetLastDate]('2016-02-03 12:34:12')
select [dbo].[fn_GetLastDate](GETDATE())

Based on the most voted answer at below link I came up with the following solution:
declare #mydate date= '2020-11-09';
SELECT DATEADD(month, DATEDIFF(month, 0, #mydate)+1, -1) AS lastOfMonth
link: How can I select the first day of a month in SQL?

I couldn't find an answer that worked in regular SQL, so I brute forced an answer:
SELECT *
FROM orders o
WHERE (MONTH(o.OrderDate) IN ('01','03','05','07','08','10','12') AND DAY(o.OrderDate) = '31')
OR (MONTH(o.OrderDate) IN ('04','06','09','11') AND DAY(o.OrderDate) = '30')
OR (MONTH(o.OrderDate) IN ('02') AND DAY(o.OrderDate) = '28')

---Start/End of previous Month
Declare #StartDate datetime, #EndDate datetime
set #StartDate = DATEADD(month, DATEDIFF(month, 0, GETDATE())-1,0)
set #EndDate = EOMONTH (DATEADD(month, DATEDIFF(month, 0, GETDATE())-1,0))
SELECT #StartDate,#EndDate