Yes, I did benchmark your code Razvan. Did you try your test-code yourself?

Running your test-code above gives 763 milliseconds for my function and 1,563 milliseconds for your function.

Mine is faster and...

The best way would be to use this function

CREATE FUNCTION dbo.GetEasterSunday 

( 

    @Y INT 

) 

RETURNS SMALLDATETIME 

AS 

BEGIN 

    DECLARE     @EpactCalc INT,  

                @PaschalDaysCalc INT, 

   ...

Right on spot, if not the case where original poster actually has an employee with id 0.

Change code above to

SELECT     a.ApplicantName,

           a.Specialization

FROM       @Applicant a

LEFT JOIN  @Shortlist s...

Ok, ok ok

Still don't get proper result with date range 1899-12-29 and 1900-01-03.

Running

select datediff(d, '19000101', '1900-01-03')/7 - datediff(d, '19000102', '1899-12-29')/7

gives...

Yes, it works well if not more than one of the two dates already is a monday. Using this code

select (datediff(d, '20060213', '20060724') + 1)/7 - datediff(d, '20060213',...

The formulas used in this topic assumes the Earth is a perfect sphere, which it is not.

The average radius of 3,958.755 miles is between the semi-major axis around the equator...

True, and it works too. But maybe not that versatile?

There is really no need for calculating the distance with the ACOS/ATAN formula, since the Earth's surface doesn't curve that much for 1.5 miles (2.4 km). Pythagoras theorem will suffice...

Another possibility is that you create a cartesian product (CROSS JOIN) between the two tables.

Something like

SELECT      Categories.Category,

            Groups.Group,

            Groups.Block

FROM        Categories

CROSS JOIN  Groups

OR

SELECT      DISTINCT Categories.Category,

            Groups.Group,

            Groups.Block

FROM        Categories

CROSS JOIN ...

Here is a shorter version

-- Do the work

SELECT      e.Employee,

            c.Code,

            c.Description,

            c.Required,

            ISNULL(h.Hours, 0) Hours

FROM        @Employees e

CROSS JOIN  @Codes c

LEFT JOIN   @Hours h ON h.Code_ID = c.ID...

-- Find duplicate data in history table

SELECT    Terc,

          Val,

          COUNT(*)

FROM      Historic

GROUP BY  Terc,

          Val

HAVING    COUNT(*) > 1

-- Find duplicate values between the two tables

SELECT     ...

-- Prepare test data

declare @employees table (id tinyint, employee char(4))

insert @employees

select 1, 'Jack' union all

select 2, 'Jill'

declare @codes table (id tinyint, code smallint, description varchar(5),...

Yes, I tried your code. I am using SQL Server 2000 with SP4. I copied all your code and pasted it into QA and ran it. The output above is what...

More efficient?

I doubt that, since you concatenate and accumulated update all rows in @table, even the duplicates. Imagine you have thousands of UserIDs where all of them have a small number...

-- prepare test data

declare @table table (UserID tinyint, usertext char(3))

insert @table

select 1, 'ABC' union all

select 1, 'DEF' union all

select 2, 'ABC' union all

select 3, 'ABC' union...