displaying dates of entire week starting day as saturday - sql-server-2008

Question

I have got the code in this site for all weeks in a year as follows,i should populate week dates, starting date as saturday and end date as friday. when the week is finished it should enter to the next week with dates.
how could i achieve this please help me.
DECLARE #Year INT=2013;
DECLARE #start DATE;
--DECLARE #WK INT=2
SET #start = DATEADD(YEAR, #Year-1900, 0);
;WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, #start)
FROM sys.all_objects
),
x AS
(
SELECT md = MIN(TDate) FROM n
WHERE DATEPART(WEEKDAY, TDate) = 7 -- assuming DATEFIRST is SATURDAY
),
y(TDate,wk) AS
(
SELECT n.TDate, ((DATEPART(DAYOFYEAR,n.TDate)-
DATEDIFF(DAY, #start,x.md)-1)/7)+1
FROM n CROSS JOIN x
WHERE n.TDate >= x.md
AND n.TDate < DATEADD(YEAR, 1, #start)
)
SELECT [date] = TDate, [week] = wk
FROM y WHERE wk < 53
ORDER BY [date];

Answer 1

Not really sure what you're asking, but based on your query above this will give week numbers, based on Saturday being the first day of the week, for 2013:
DECLARE #Year INT=2013;
DECLARE #start DATE;
SET #start = DATEADD(YEAR, #Year-1900, 0);
SET DATEFIRST 6; -- Set start of week as Saturday
WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, #start)
FROM sys.all_objects
)
select TDate
, DATEPART(WEEK,TDate)
from n
where year(TDate) = 2013;
Edit:
So based on the various comments and answers I think what is required here is, for a given day, return all days in that same week, with Saturday as the first day of the week. So something like this:
set datefirst 6; -- make sure first day of week is Saturday
declare #date date = getdate(); -- change date as required here
with daysOfWeek as
(
select [date] = dateadd(dd, -datepart(dw, #date) + 1, #date)
union all
select [date] = dateadd(dd, 1, [date])
from daysOfWeek
where [date] < dateadd(dd, -datepart(dw, #date) + 7, #date)
)
select [date], dayOfWeek = datename(dw, [date])
from daysOfWeek
Which gives results:
I think this is what's required here?
Second edit:
First, create the function:
create function dbo.weekDates (#date date)
returns #dates table ([date] date, [dayofweek] varchar(9))
as
begin
with daysOfWeek as
(
select [date] = dateadd(dd, -datepart(dw, #date) + 1, #date)
union all
select [date] = dateadd(dd, 1, [date])
from daysOfWeek
where [date] < dateadd(dd, -datepart(dw, #date) + 7, #date)
)
insert into #dates ([date], [dayofweek])
select [date], [dayOfWeek] = datename(dw, [date])
from daysOfWeek;
return
end
go
Using the function:
set datefirst 6 -- Set Saturday as first day of week
select * from dbo.weekDates (getdate()) -- Change input parameter as required

Answer 2

I HAVE ACHIEVED TO DISPLAY THE WEEK DATES BUT IT IS DISPLAYING THE NEXT WEEK DATES I WANT TO DISPLAY THE CURRENT WEEK DATES, HERE START DAY IS SATURDAY AND END DAY IS FRIDAY
ALTER FUNCTION GetCurrentWeek()
RETURNS #TWeek TABLE (TWeek NVARCHAR(20))
AS
BEGIN
DECLARE #Year INT= DATEPART(YEAR,GETDATE());
DECLARE #start DATE;
SET #start = DATEADD(YEAR, #Year-1900, 0);
;WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, #start)
FROM sys.all_objects
),
x AS
(
SELECT md = MIN(TDate) FROM n
WHERE DATEPART(WEEKDAY, TDate) = 7 -- assuming DATEFIRST is SATURDAY
),
y(TDate,wk) AS
(
SELECT n.TDate, ((DATEPART(DAYOFYEAR, n.TDate)
- DATEDIFF(DAY, #start, x.md)-1)/7) + 1
FROM n CROSS JOIN x
WHERE n.TDate >= x.md
AND n.TDate < DATEADD(YEAR, 1, #start)
)
INSERT #TWeek
SELECT [date] = TDate
FROM y WHERE wk =DATEPART(wk, GetDate())
ORDER BY [date];
RETURN;
END