returns the current value, for a session, of SET DATE FIRST, i.e it returns first day of week, TinyInt Value from 1 through 7 representing Monday to Sunday as shown Below.

1 for Monday
2 for Tuesday
3 for Wednesday
4 for Thursday
5 for Friday
6 for Saturday
7 for Sunday

For Example, If the first day of week is Sunday (US) , Then @@DateFirst would return 7.

If the first day of week is Monday (Europe), Then @@DateFirst Would return 1.

SET DateFirst = 3

The above statement would set Wednesday as first day of the week for the current Session.

‘July 1st 2009 is a wednesday

SET DateFirst = 1 -- or SET LANGUAGE Italian;
Select DATEPART(dw, '20090701')

would return 3 (since Wednesday is 3rd day from Monday)

SET DateFirst = 7 -- or SET LANGUAGE us_english;
Select DATEPART(dw, '20090701')

would return 4 (since Wednesday is 4th day from Sunday)

DATEPART return the weekday according @@DateFirst.

CREATE FUNCTION dbo.GetBusinessDays
(
@startDate SMALLDATETIME,
@endDate SMALLDATETIME
)
RETURNS INT
AS
BEGIN
DECLARE @dateDiff INT;

SET @dateDiff = DATEDIFF(DAY, @startDate, @endDate)+1;

RETURN
(
SELECT @dateDiff/ 7 * 5 + CASE WHEN (@dateDiff % 7) = 0 THEN 0

ELSE

CASE
When (DATEPART(WEEKDAY, @endDate ) + @@DateFirst -1)%7 = 0 and (@dateDiff % 7) =1 Then (@dateDiff % 7)-1 –If it is sunday

When (DATEPART(WEEKDAY, @endDate ) + @@DateFirst -1)%7 = 0 and (@dateDiff % 7) <>1 Then (@dateDiff % 7)-2 –If it is sunday

When (DATEPART(WEEKDAY, @endDate ) + @@DateFirst -1)%7 =6 Then (@dateDiff % 7) – 1   –If it is Saturday

ELSE (@dateDiff % 7)
END — Inner Case

END

);
END

There are many other ways of achieving this , you can also look at the following examples

CREATE FUNCTION dbo.GetWorkingDays
(
@startDate SMALLDATETIME,
@endDate SMALLDATETIME
)
RETURNS INT
AS
BEGIN
DECLARE @range INT;

SET @range = DATEDIFF(DAY, @startDate, @endDate)+1;

RETURN
(
SELECT
@range / 7 * 5 + @range % 7 -
(
SELECT COUNT(*)
FROM
(
SELECT 1 AS d
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
) weekdays
WHERE d <= @range % 7
AND DATENAME(WEEKDAY, @endDate – d + 1)
IN
(
‘Saturday’,
‘Sunday’
)
)
);
END
GO

or this that i found on one of the forums

DECLARE @start_date DATETIME;
DECLARE @end_date DATETIME;

SET @start_date = '20080820';
SET @end_date = '20080825';

SELECT ((total_days / 7) * 5 + total_days % 7 -
CASE WHEN 6 BETWEEN start_weekday AND end_weekday
THEN 1 ELSE 0 END -
CASE WHEN 7 BETWEEN start_weekday AND end_weekday
THEN 1 ELSE 0 END)
FROM (SELECT total_days, start_weekday,
start_weekday + total_days % 7 - 1
FROM (SELECT DATEDIFF(day, @start_date, @end_date) + 1,
DATEPART(WEEKDAY, @start_date + @@DATEFIRST - 1)
) AS T(total_days, start_weekday)
) AS D(total_days, start_weekday, end_weekday);

NOTE : This example doesnot holidays other than saturday and sunday into account.

http://msdn.microsoft.com/library/default.asp?url=/library/en-us/optimsql/odp_tun_1a_6oxf.asp

It's slightly out-of-date, and I'll work on trying to get it updated.  Specifically, SQL 2005 will remote uniqueidentifiers (and it will support statistics over them as well).  We'll remote queries from SQL 2005 to SQL 2000 that contain filters comparing columns to constant uniqueidentifier values as well.

We published a research paper last year on how the Distributed Query feature works in more detail.  While it does not cover every practical detail of the implementation, you may find it as an interesting reference if you use distributed queries.

http://citeseer.ist.psu.edu/732761.html

If you have other remoting questions/problems, please post comments on them and we'll see if we can get them answered for you.

Thanks,

Conor

Source : http://blogs.msdn.com/queryoptteam/default.aspx