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How to get max Saturday dates in a column of each month in SQL Server. Can someone please help me.
Now I need only the dates which has last Saturday of month.
For example,
The table has
07-08-2021 - Saturday
14-08-2021 - Saturday
21-08-2021 - Saturday
28-08-2021 - Saturday
04-09-2021 - Saturday
11-09-2021 - Saturday
18-09-2021 - Saturday
25-09-2021 - Saturday
Suppose we are in August month, I need to select last Saturday of that month( ONLY 28-08-2021)
Suppose we are in September month, I need to select last Saturday of that month( ONLY 25-09-2021)
Output:
28-08-2021
25-09-2021
assuming you have a datefield in your table (I will refer to it here as such in the below query)
with week as (
select
date_trunc('week', datefield + interval '2 day') - interval '2 day' as week_date
-- ^this adjusts the week date_trunc to start on Saturday (it starts on Monday by default)
from sometable
)
select
extract(month from week_date) as month_num,
max(week_date) as last_saturday
from week
group by month_num
note: if you only have partial data for the current month, this query will need to be altered slightly, but you didn't give me a lot to go off of here
A CTE is defined to populate the day name of entire month and then the requirement is filtered.
;with GetDates As (
select CAST('01-01-2021' as date) as StartDate, datename(dw, '09-01-2021') as Day_Name
UNION ALL
select DATEADD(day,1, StartDate), datename(dw, DATEADD(day,1, StartDate))
from GetDates
where StartDate < '09-30-2021'
)
select max(StartDate)
from GetDates where Day_Name = 'Saturday'
group by month(StartDate)
OPTION (MAXRECURSION 500)
Here is a function to calculate the Nth Weekday - which can calculate from the beginning of the month or from the end of the month.
Alter Function dbo.fnGetNthWeekDay (
#theDate datetime
, #theWeekday int
, #theNthDay int
)
Returns Table
As
Return
/*
Adapted from a version published by Peter Larrson - with minor modifications for performance
and restructured to eliminate usage of a derived table.
Inputs:
#theDate any date in the month
#theWeekday the weekday to calculate: 1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
7 = Sunday
#theNthDay the week count where positive is from beginning of the month
and negative is from end of month
Outputs:
#theDate the date entered
#theNthDate the Nth date of the month
*/
Select theDate = #theDate
, dt.nthDate
From (Values (dateadd(month, datediff(month, #theNthDay, #theDate), 0))) As mm(FirstOfMonth)
Cross Apply (Values (dateadd(day, 7 * #theNthDay - 7 * sign(#theNthDay + 1)
+ (#theWeekday + 6 - datediff(day, -53690, mm.FirstOfMonth) % 7) % 7, mm.FirstOfMonth))) As dt(nthDate)
Where #theWeekday Between 1 And 7
And datediff(month, dt.nthDate, #theDate) = 0
And #theNthDay In (-5, -4, -3, -2, -1, 1, 2, 3, 4, 5);
Go
You can then call it like this:
Select * From dbo.fnGetNthWeekDay('2021-08-15', 6, -1) nwd
How do I get yesterdays date and if the current date is Monday I would need Sunday, Saturday and Friday. This is already asked here for ms access. I now need this for SQL server. How would I go about this?
Will create an inline view that will return the previous date in SQL It will return between 1 and 3 rows depending on the current date.
If current date is Monday Return:
Sundays date
Saturdays date
Fridays date
If current date is Tuesday then Return: Mondays date
If current date is Wednesday then Return: Tuesdays date
If current date is Thursday then Return: Wednesdays date
If current date is Friday then Return: Thursdays date
If current date is Saturday then Return: Fridays date
If current date is Sunday then Return: Saturdays date
I hope this helps explain what I am trying to do more clearly.
sample select query
--get previous date
select * from [Purchase Orders] where MyDate in (previous date(s))
Not entirely clear about the question though. But you can do something on the similar lines.
Using DATEPART and DATEADD
DECLARE #TodaysDate DATETIME
SET #TodaysDate = '2013-04-10'
SELECT CASE
WHEN DATEPART (DW, DATEADD (DD, -1, #TodaysDate )) IN (1, 6, 7)
THEN 'WeekEnd' ELSE 'WeekDay'
END D
Using the query you linked to, you can try the code below:
SELECT *
FROM [Purchase Order]
WHERE MyDate >= CASE
WHEN DATENAME(dw, CONVERT(CHAR(8), GETDATE() , 112)) LIKE 'Monday' THEN CONVERT(CHAR(8), DATEADD(dd, -3, GETDATE()), 112)
ELSE CONVERT(CHAR(8), DATEADD(dd, -1, GETDATE()), 112)
END
AND MyDate < CONVERT(CHAR(8),GETDATE(),112)
You can use the following table function:
CREATE FUNCTION dbo.ufnPreviousDay (#suppliedDate DATE)
RETURNS #PreviousDates TABLE (PreviousDate DATE)
AS
BEGIN
INSERT INTO #PreviousDates (
PreviousDate)
SELECT
PreviousDate = DATEADD(DAY, -1, #suppliedDate)
IF DATEPART(WEEKDAY, DATEADD(DAY, ##DATEFIRST - 1, #suppliedDate)) = 1 -- If Monday
BEGIN
INSERT INTO #PreviousDates (
PreviousDate)
SELECT
PreviousDate = DATEADD(DAY, -2, #suppliedDate)
UNION ALL
SELECT
PreviousDate = DATEADD(DAY, -3, #suppliedDate)
END
RETURN
END
You can use it like this:
SELECT
T.*
FROM
dbo.ufnPreviousDay('2018-03-19') AS T
/*
Result:
2018-03-18
2018-03-17
2018-03-16
*/
If you want today's previous dates:
SELECT
T.*
FROM
YourTable AS T
INNER JOIN dbo.ufnPreviousDay(GETDATE()) AS M ON T.MyDate = M.PreviousDate
I tried but could not get the right solution. I want an SQL query that lists all the weekend dates of the current year.
I tried this SQL query:
WITH hier(num, lvl) AS (
SELECT 0, 1
UNION ALL
SELECT 100, 1
UNION ALL
SELECT num + 1, lvl + 1
FROM hier
WHERE lvl < 100
)
SELECT lvl [Week],
convert(date,DATEADD(dw, -DATEPART(dw, DATEADD(wk,DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ),
DATEADD(wk, DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ) - num * 7,101) [End Date]
FROM hier a
where num < 52
ORDER BY [End Date] asc
Its output is like this:
Week End date
52 2012-01-14
51 2012-01-21
50 2012-01-28
49 2012-02-04
I want the dates to start from the beginning – so, the above is missing one weekend, which is 2012-07-01. Also, I want the week numbers to show as 1, 2, 3... instead of 52, 51....
Check out this blog post.
Your question is explained in detail.
DECLARE #Year AS INT,
#FirstDateOfYear DATETIME,
#LastDateOfYear DATETIME
-- You can change #year to any year you desire
SELECT #year = 2010
SELECT #FirstDateOfYear = DATEADD(yyyy, #Year - 1900, 0)
SELECT #LastDateOfYear = DATEADD(yyyy, #Year - 1900 + 1, 0)
-- Creating Query to Prepare Year Data
;WITH cte AS (
SELECT 1 AS DayID,
#FirstDateOfYear AS FromDate,
DATENAME(dw, #FirstDateOfYear) AS Dayname
UNION ALL
SELECT cte.DayID + 1 AS DayID,
DATEADD(d, 1 ,cte.FromDate),
DATENAME(dw, DATEADD(d, 1 ,cte.FromDate)) AS Dayname
FROM cte
WHERE DATEADD(d,1,cte.FromDate) < #LastDateOfYear
)
SELECT FromDate AS Date, Dayname
FROM CTE
WHERE DayName IN ('Saturday','Sunday') -- For Weekend
/*
WHERE DayName LIKE 'Sunday'
WHERE DayName NOT IN ('Saturday','Sunday') -- For Weekday
WHERE DayName LIKE 'Monday' -- For Monday
WHERE DayName LIKE 'Sunday' -- For Sunday
*/
OPTION (MaxRecursion 370)
Will this help
DECLARE #startDate DATETIME, #endDate DATETIME
SELECT #startDate = '2012-01-01', #endDate = '2012-12-31'
;WITH Calender AS (
SELECT #startDate AS dt
UNION ALL
SELECT dt + 1 FROM Calender
WHERE dt + 1 <= #endDate
)
SELECT
dt
,NameMonth = DATENAME(Month, dt)
,NameDay = DATENAME (Weekday,dt)
,WeekofYr = DATEPART(WEEK, dt) FROM Calender
WHERE DATENAME (Weekday,dt) IN ('Sunday')
Option(MaxRecursion 0)
Result(Partial)
dt NameMonth NameDay WeekofYr
2012-01-01 00:00:00.000 January Sunday 1
2012-01-08 00:00:00.000 January Sunday 2
...............................................
...............................................
2012-12-30 00:00:00.000 December Sunday 53
you can try this
DECLARE #FirstDateOfYear DATETIME
SET #FirstDateOfYear = ’2010-01-01′
SELECT DISTINCT DATEADD(d, number, #FirstDateOfYear),
CASE DATEPART(dw, DATEADD(d, number, #FirstDateOfYear))
WHEN 7 THEN ‘Saturday’
WHEN 1 THEN ‘Sunday’
ELSE ‘Work Day’
END
FROM master..spt_values
WHERE number BETWEEN 0 AND 364
AND (DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 1 OR DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 7)
ORDER BY DATEADD(d, number, #FirstDateOfYear)
Try to find the first Saturday by doing this:
Start on 2012-01-01
If it's not a Saturday, add a day
Goto 2
Then, into a temporary table, add that date and the following date (Sunday).
After that, loop the following:
Add 7 and 8 days to the last Saturday you found (you get the following Saturday and Sunday)
Check whether they are still in 2012
If they are, store them in temp table and goto 1
There may be more elegant ways, but that's my quick & dirty solution. As you didn't post any code of what you've tried, I'll leave the implementation up to you.
this also works
declare #dat datetime, #add int
set #dat = '20120101'
set #add = datepart(w,#dat)
set #add = 5 - #add -- friday
set #dat = dateadd(d,#add,#dat)
while #dat <= '20121231'
begin
print #dat
set #dat = dateadd(d,7,#dat)
end
;with AllDaysOfYear (Day) as (
select DATEADD(year,DATEDIFF(year,0,CURRENT_TIMESTAMP),0) --Jan 1st
union all
select DATEADD(day,1,Day) from AllDaysOfYear
where DATEPART(year,DATEADD(day,1,Day)) = DATEPART(year,CURRENT_TIMESTAMP)
)
select
ROW_NUMBER() OVER (ORDER BY Day) as WeekNo,
Day
from
AllDaysOfYear
where
DATEPART(weekday,Day) = DATEPART(weekday,'20120714')
option (maxrecursion 0)
First, generate a set of all of the days in the current year (AllDaysInYear). Then, select those whose weekday is a saturday. The value I've used ('20120714') isn't terribly important - it just has to be any saturday, from any year. I'm just using it to avoid needing to have particular DATEFIRST or language settings.
This query shows how to get the first day of this year and the first day of the next year in the first part. The first day of the next year is calculated once so as not to keep getting and comparing the year parts.
;WITH cte(TheDate,NextYear) AS
(
SELECT CAST(CONVERT(CHAR(4),GETDATE(),112)+'0101' AS DATETIME),
CAST(YEAR(GETDATE())*10000+10101 AS CHAR(8))
UNION ALL
SELECT DateAdd(d,1,TheDate),NextYear
FROM cte
WHERE DateAdd(d,1,TheDate)<NextYear
)
SELECT Week = DatePart(wk,TheDate),
TheDate
FROM cte
WHERE DateName(dw,TheDate) in ('Saturday')
ORDER BY TheDate
OPTION (MAXRECURSION 366)
with t as
(
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
)
select * from
(
select
current_timestamp
-datepart(dy,current_timestamp)
+row_number() over (order by t.b) d
from t, t t1, t t2
) tmp
where datepart(yyyy,d)=datepart(yyyy,current_timestamp)
and
DATENAME(dw,d)='sunday'
DECLARE #Year AS INT
SELECT #Year = 2020
;WITH weekends AS (
SELECT DATEFROMPARTS(#Year, 1, 1) AS dt
UNION ALL
SELECT DATEADD(DAY, 1, dt)
FROM weekends
WHERE dt < DATEFROMPARTS(#Year, 12, 31)
)
SELECT dt, DATENAME(MONTH, dt), DATENAME(DW, dt)
FROM weekends
WHERE DATEPART(DW, dt) IN (1, 7)
OPTION(MaxRecursion 366)
Is there is a way to group dates by week of month in SQL Server?
For example
Week 2: 05/07/2012 - 05/13/2012
Week 3: 05/14/2012 - 05/20/2012
but with Sql server statement
I tried
SELECT SOMETHING,
datediff(wk, convert(varchar(6), getdate(), 112) + '01', getdate()) + 1 AS TIME_
FROM STATISTICS_
GROUP BY something, TIME_
ORDER BY TIME_
but it returns the week number of month. (means 3)
How to get the pair of days for current week ?
For example, now we are in third (3rd) week and I want to show 05/14/2012 - 05/20/2012
I solved somehow:
SELECT DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)
select DATEADD(ww, DATEDIFF(ww,0,<my_column_name>), 0)+6
Then I will get two days and I will concatenate them later.
All right, bear with me here. We're going to build a temporary calendar table that represents this month, including the days from before and after the month that fall into your definition of a week (Monday - Sunday). I do this in a lot of steps to try to make the process clear, but I probably haven't excelled at that in this case.
We can then generate the ranges for the different weeks, and you can join against your other tables using that.
SET DATEFIRST 7;
SET NOCOUNT ON;
DECLARE #today SMALLDATETIME, #fd SMALLDATETIME, #rc INT;
SELECT #today = DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0), -- today
#fd = DATEADD(DAY, 1-DAY(#today), #today), -- first day of this month
#rc = DATEPART(DAY, DATEADD(DAY, -1, DATEADD(MONTH, 1, #fd)));-- days in month
DECLARE #thismonth TABLE (
[date] SMALLDATETIME,
[weekday] TINYINT,
[weeknumber] TINYINT
);
;WITH n(d) AS (
SELECT TOP (#rc+12) DATEADD(DAY, ROW_NUMBER() OVER
(ORDER BY [object_id]) - 7, #fd) FROM sys.all_objects
)
INSERT #thismonth([date], [weekday]) SELECT d, DATEPART(WEEKDAY, d) FROM n;
DELETE #thismonth WHERE [date] < (SELECT MIN([date]) FROM #thismonth WHERE [weekday] = 2)
OR [date] > (SELECT MAX([date]) FROM #thismonth WHERE [weekday] = 1);
;WITH x AS ( SELECT [date], weeknumber, rn = ((ROW_NUMBER() OVER
(ORDER BY [date])-1) / 7) + 1 FROM #thismonth ) UPDATE x SET weeknumber = rn;
-- now, the final query given all that (I've only broken this up to get rid of the vertical scrollbars):
;WITH ranges(w,s,e) AS (
SELECT weeknumber, MIN([date]), MAX([date]) FROM #thismonth GROUP BY weeknumber
)
SELECT [week] = CONVERT(CHAR(10), r.s, 120) + ' - ' + CONVERT(CHAR(10), r.e, 120)
--, SOMETHING , other columns from STATISTICS_?
FROM ranges AS r
-- LEFT OUTER JOIN dbo.STATISTICS_ AS s
-- ON s.TIME_ >= r.s AND s.TIME_ < DATEADD(DAY, 1, r.e)
-- comment this out if you want all the weeks from this month:
WHERE w = (SELECT weeknumber FROM #thismonth WHERE [date] = #today)
GROUP BY r.s, r.e --, SOMETHING
ORDER BY [week];
Results with WHERE clause:
week
-----------------------
2012-05-14 - 2012-05-20
Results without WHERE clause:
week
-----------------------
2012-04-30 - 2012-05-06
2012-05-07 - 2012-05-13
2012-05-14 - 2012-05-20
2012-05-21 - 2012-05-27
2012-05-28 - 2012-06-03
Note that I chose YYYY-MM-DD on purpose. You should avoid regional formatting like M/D/Y especially for input but also for display. No matter how targeted you think your audience is, you're always going to have someone who thinks 05/07/2012 is July 5th, not May 7th. With YYYY-MM-DD there is no ambiguity whatsoever.
Create a calendar table, then you can query week numbers, first/last days of specific weeks and months etc. You can also join on it queries to get a date range etc.
How about a case statement?
case when datepart(day, mydatetime) between 1 and 7 then 1
when datepart(day, mydatetime) between 8 and 14 then 2
...
You'll also have to include the year & month unless you want all the week 1s in the same group.
It's not clear of you want to "group dates by week of month", or alternately "select data from a given week"
If you mean "group" this little snippet should get you 'week of month':
SELECT <stuff>
FROM CP_STATISTICS
WHERE Month(<YOUR DATE COL>) = 5 --april
GROUP BY Year(<YOUR DATE COL>),
Month(<YOUR DATE COL>),
DATEDIFF(week, DATEADD(MONTH, DATEDIFF(MONTH, 0, <YOUR DATE COL>), 0)
, <YOUR DATE COL>) +1
Alternately, if you want "sales for week 1 of April, ordered by date" You could do something like..
DECLARE #targetDate datetime2 = '5/3/2012'
DECLARE #targetWeek int = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, #targetDate), 0), #targetDate) +1
SELECT <stuff>
FROM CP_STATISTICS
WHERE MONTH(#targetDate) = Month(myDateCol) AND
YEAR(#targetDate) = Year (myDateCol) AND
#targetWeek = DATEDIFF(week, DATEADD(MONTH,
DATEDIFF(MONTH, 0, myDateCol), 0), myDateCol) +1
ORDER BY myDateCol
Note, things would get more complicated if you use non-standard weeks, or want to reach a few days into an earlier month for weeks that straddle a month boundary.
EDIT 2
From looking at your 'solved now' section. I think your question is "how do I get data out of a table for a given week?"
Your solution appears to be:
DECLARE #targetDate datetime2 = '5/1/2012'
DECLARE #startDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,targetDate), 0)
DECLARE #endDate datetime2 = DATEADD(ww, DATEDIFF(ww,0,#now), 0)+6
SELECT <stuff>
FROM STATISTICS_
WHERE dateStamp >= #startDate AND dateStamp <= #endDate
Notice how if the date is 5/1 this solution results in a start date of '4/30/2012'. I point this out because your solution crosses month boundaries. This may or may not be desirable.
I have a table called FcData and the data looks like:
Op_Date
2011-02-14 11:53:40.000
2011-02-17 16:02:19.000
2010-02-14 12:53:40.000
2010-02-17 14:02:19.000
I am looking to get the Number of weeks in That Month from Op_Date. So I am looking for output like:
Op_Date Number of Weeks
2011-02-14 11:53:40.000 5
2011-02-17 16:02:19.000 5
2010-02-14 12:53:40.000 5
2010-02-17 14:02:19.000 5
This page has some good functions to figure out the last day of any given month: http://www.sql-server-helper.com/functions/get-last-day-of-month.aspx
Just wrap the output of that function with a DATEPART(wk, last_day_of_month) call. Combining it with an equivalent call for the 1st-day-of-week will let you get the number of weeks in that month.
Use this to get the number of week for ONE specific date. Replace GetDate() by your date:
declare #dt date = cast(GetDate() as date);
declare #dtstart date = DATEADD(day, -DATEPART(day, #dt) + 1, #dt);
declare #dtend date = dateadd(DAY, -1, DATEADD(MONTH, 1, #dtstart));
WITH dates AS (
SELECT #dtstart ADate
UNION ALL
SELECT DATEADD(day, 1, t.ADate)
FROM dates t
WHERE DATEADD(day, 1, t.ADate) <= #dtend
)
SELECT top 1 DatePart(WEEKDAY, ADate) weekday, COUNT(*) weeks
FROM dates d
group by DatePart(WEEKDAY, ADate)
order by 2 desc
Explained: the CTE creates a result set with all dates for the month of the given date. Then we query the result set, grouping by week day and count the number of occurrences. The max number will give us how many weeks the month overlaps (premise: if the month has 5 Mondays, it will cover five weeks of the year).
Update
Now, if you have multiple dates, you should tweak accordingly, joining your query with the dates CTE.
Here is my take on it, might have missed something.
In Linq:
from u in TblUsers
let date = u.CreateDate.Value
let firstDay = new DateTime(date.Year, date.Month, 1)
let lastDay = firstDay.AddMonths(1)
where u.CreateDate.HasValue
select Math.Ceiling((lastDay - firstDay).TotalDays / 7)
And generated SQL:
-- Region Parameters
DECLARE #p0 Int = 1
DECLARE #p1 Int = 1
DECLARE #p2 Float = 7
-- EndRegion
SELECT CEILING(((CONVERT(Float,CONVERT(BigInt,(((CONVERT(BigInt,DATEDIFF(DAY, [t3].[value], [t3].[value2]))) * 86400000) + DATEDIFF(MILLISECOND, DATEADD(DAY, DATEDIFF(DAY, [t3].[value], [t3].[value2]), [t3].[value]), [t3].[value2])) * 10000))) / 864000000000) / #p2) AS [value]
FROM (
SELECT [t2].[createDate], [t2].[value], DATEADD(MONTH, #p1, [t2].[value]) AS [value2]
FROM (
SELECT [t1].[createDate], CONVERT(DATETIME, CONVERT(NCHAR(2), DATEPART(Month, [t1].[value])) + ('/' + (CONVERT(NCHAR(2), #p0) + ('/' + CONVERT(NCHAR(4), DATEPART(Year, [t1].[value]))))), 101) AS [value]
FROM (
SELECT [t0].[createDate], [t0].[createDate] AS [value]
FROM [tblUser] AS [t0]
) AS [t1]
) AS [t2]
) AS [t3]
WHERE [t3].[createDate] IS NOT NULL
According to this MSDN article: http://msdn.microsoft.com/en-us/library/ms174420.aspx you can only get the current week in the year, not what that month returns.
There may be various approaches to implementing the idea suggested by #Marc B. Here's one, where no UDFs are used but the first and the last days of month are calculated directly:
WITH SampleData AS (
SELECT CAST('20110214' AS datetime) AS Op_Date
UNION ALL SELECT '20110217'
UNION ALL SELECT '20100214'
UNION ALL SELECT '20100217'
UNION ALL SELECT '20090214'
UNION ALL SELECT '20090217'
),
MonthStarts AS (
SELECT
Op_Date,
MonthStart = DATEADD(DAY, 1 - DAY(Op_Date), Op_Date)
/* alternatively: DATEADD(MONTH, DATEDIFF(MONTH, 0, Op_Date), 0) */
FROM FcData
),
Months AS (
SELECT
Op_Date,
MonthStart,
MonthEnd = DATEADD(DAY, -1, DATEADD(MONTH, 1, MonthStart))
FROM FcData
)
Weeks AS (
SELECT
Op_Date,
StartWeek = DATEPART(WEEK, MonthStart),
EndWeek = DATEPART(WEEK, MonthEnd)
FROM MonthStarts
)
SELECT
Op_Date,
NumberOfWeeks = EndWeek - StartWeek + 1
FROM Weeks
All calculations could be done in one SELECT, but I chose to split them into steps and place every step in a separate CTE so it could be seen better how the end result was obtained.
You can get number of weeks per month using the following method.
Datepart(WEEK,
DATEADD(DAY,
-1,
DATEADD(MONTH,
1,
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())))
-
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())
+1
)
Here how you can get accurate amount of weeks:
DECLARE #date DATETIME
SET #date = GETDATE()
SELECT ROUND(cast(datediff(day, dateadd(day, 1-day(#date), #date), dateadd(month, 1, dateadd(day, 1-day(#date), #date))) AS FLOAT) / 7, 2)
With this code for Sep 2014 you'll get 4.29 which is actually true since there're 4 full weeks and 2 more days.