The first day of the week

Under some assumptions the task can be solved as follows:

WITH num(n) AS(/* With aid of [recursive CTE](/ru/book_recursive_cte.html "Общие табличные выражения") get a table having single column n
                     with sequence of integer values from 0 to 6 */
SELECT 0
UNION ALL
SELECT n+1 FROM num
WHERE n < 6),
dat AS (/* Getting table with dates from January 1 to 7, 2013  */
SELECT DATEADD(dd,  n,  CAST('2013-01-01' AS DATETIME)) AS day FROM num
)
SELECT day FROM dat  WHERE DATEPART(dw, day) = 1; /* Taking the day that corresponds
                        to the first day of a week */

The assumption we mentioned above adopts Monday as the first day of a week. However, running of the query on the sql-ex.ru site gives us

WITH num(n) AS (SELECT 0
UNION ALL
SELECT n+1 FROM num
WHERE n < 6),
dat AS (
SELECT DATEADD(dd,  n,  CAST('2013-01-01' AS DATETIME)) AS day FROM num
)
SELECT day FROM dat  WHERE DATENAME(dw, day) = 'monday';

This is true now, but will this query be true always if we shall embed it into application code? The answer is no. If language settings will be changed

SET
LANGUAGE german;

we’ll get empty row set because the last line in the query should be written as

SELECT day FROM dat  WHERE DATENAME(dw, day) ='Montag';

@@DATEFIRST function

@@DATEFIRST returns an integer which defines the first day of a week for the current session. In so doing, 1 is corresponding to Monday, but 7 - to Sunday respectively. I.e. if

SELECT
@@DATEFIRST;

To make our solution to the task be independent on the first-day-of-week settings, we need to use @@DATEFIRST function, for example

WITH num(n) AS(
SELECT 0
UNION ALL
SELECT n+1 FROM num
WHERE n < 6),
dat AS (
SELECT DATEADD(dd,  n,  CAST('2013-01-01' AS DATETIME)) AS day FROM num
)
SELECT day, DATENAME(dw, day) week_day FROM dat  WHERE DATEPART(dw, day) =
1+(8-@@DATEFIRST) % 7;

To change value of the first day of a week (for a current session), we can use SET DATEFIRST statement. So, if we run the following script

SET DATEFIRST 1;
SELECT @@DATEFIRST;

we’ll obtain 1, i.e. week starts with Monday now.

The first-day-of-a-week settings may be changed through the changing of session language. For example, running 

SET LANGUAGE us_english;
SELECT
@@DATEFIRST;

lead us to 7 (Sunday) once more, as this value is default for U.S. English, whereas choosing German will do Monday to be the first day of a week.

Nevertheless you can change the language parameter and first day of a week simultaneously to obtain, say, English and a week that will start with Monday:

SET DATEFIRST 1;
SET LANGUAGE us_english;
SELECT @@DATEFIRST;

Language settings impact specifically on the string representation of date/time components. Compare these results:

DECLARE @dt DATE = '2012-12-17'; -- December 17, 2012
SET LANGUAGE us_english;
SELECT DATENAME(DW, @dt) AS day_of_week, DATENAME(MONTH,@dt) AS month;

SET LANGUAGE german;
SELECT DATENAME(DW, @dt) AS day_of_week, DATENAME(MONTH, @dt) AS month;

Conclusion. If you want to have queries that always give correct results for any settings of server or database, you can specify necessary setting parameters for the session in which these queries are executing; or you need to check the values of corresponding parameters (@@DATEFIRST as example) in each such a query.

declare  @anchor_date datetime
declare  @reference_date datetime  
  
select @anchor_date='19000101', @reference_date='20130505'  
  
SELECT DATEADD(day, DATEDIFF(day, @anchor_date,  
  
DATEADD(year, DATEDIFF(year, '19000101', @reference_date), '19000101') - 1) /7*7 + 7,  
  
@anchor_date);

SELECT day FROM dat  WHERE DATEPART(dw, day) = DATEPART(dw, '19000101');

SELECT day FROM dat  WHERE DATENAME(dw, day) = DATENAME(dw, '19000101');

Seggested exercises: 78, 99, 110, 118