sql

problem:
finding the duplicate within sql

solution:
provide table named subscriber, with elements: email

SELECT COUNT(DISTINCT email) AS NumberOfEmail, email FROM subscriber GROUP BY email HAVING (COUNT(email) > 1)

problem:
make some selection in database OTHER THEN top (10) post

solution:  
provide, table named news with element: newsID, newsDate

SELECT * FROM News WHERE (newsID NOT IN (SELECT TOP (10) newsID FROM News AS News_1 ORDER BY newsDate DESC)) ORDER BY newsDate DESC 

it will return all post except top 10 post 

problem:
display age with sql command

solution:
sql command:
SELECT     DATEDIFF(mm, dateOfBirth, GETDATE()) / 12 AS age
FROM         someTable

it wil display age ^^V

problem:

displaying list of bday boy/girl in next (30/ according to your own choice ) days from the current date.

solution:
im using the sql to select the date of birth, for displaying, you can use whatever you want, like gridview, datalist etc.

hereby the sql command:

SELECT     DateOfBirth
FROM         someTablename
WHERE     (DateOfBirth < GETDATE()) AND (DATEDIFF(d, GETDATE(), DATEADD(yy, DATEDIFF(yy, DateOfBirth, GETDATE()), DateOfBirth)) BETWEEN
                      0 AND 30)

it will return the list of date of birth from the current date.

for reference of the sql command:
http://msdn.microsoft.com/en-us/library/aa258269%28SQL.80%29.aspx

SELECT previous row or "LIMIT-command" in MSSQL

SELECT TOP (1) FROM table_name WHERE (ID< @ID)
ORDER BY ID DESC

SELECT next row or "LIMIT-command" in MSSQL

SELECT TOP (1) FROM table_name WHERE (ID> @ID)
ORDER BY ID

Thank also for the descussion