does this change from SQL2000 to SQL2005/8? when i try this in my SQL2000 if i leave the % out of the select- i get zero rows. needless to say when I try the test question in my database only the ones that have an 'n' in the third position are dropped so the answer appears to be three. and according the MSDN site it says:
The following example uses the [^] operator to find all the people in the Contact table who have first names that start with Al and have a third letter that is not the letter a.
Copy Code USE AdventureWorks;
SELECT FirstName, LastName
WHERE FirstName LIKE 'Al[^a]%'
ORDER BY FirstName;
not trying to be difficult- trying to understand what i am missing.. thanks