Then the CAST solution is probably the best for you, eg:

DECLARE @Int1 INT

SET @int1 = 454638

SELECT @Int1, LEFT(CAST(@int1 AS VARCHAR(20)),3)

You have stated your problem and received a solution - if the solution is not what you require, please describe why, so that the next solution can give you what...

Although you may have solved this problem in the short term, you may be storing up problems for yourself in the medium term. If two or three more code/description pairs...

Do you have a question?

Haha, no problem. SSCrazy would frequently be a better name for me than Phil...

It is not. Your method would allow invalid ranges to be inserted, mine would not.

But...

Jayanth_Kurup (6/27/2011)

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Hi

This would remove scope for inserting a range between 1 and...

Something like this:

declare @MaxToPoint Int

select @MaxToPoint = max(ToPoint) from PLevel

If @frmPoint <= @MaxToPoint then --> error

I am assuming that selecting MAX(ToPoint) and making sure that your range is above that is not adequate?

Please mark the bit that is creating the error. 'Third step' does not make it clear to me.

Also, please post the text of the error message.

Why does it fail?

Why do you need a cursor?

It's not easy to understand what you are asking - can you provide sample input data, with desired output?

GilaMonster (6/23/2011)

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There's a massive difference between the two.

The first there's a function applied to the column. That makes the predicate non-SARGable, meaning SQL can NEVER use an index seek operation...

As you are looking from the start of the column, you can use this

WHERE LEFT(fldField,3) = 'NCB'

- just a little tidier and avoids wildcards.

Here is proof that "it works"

DECLARE @a CHAR(1) = CHAR(10)

SELECT @a, ASCII(@a), REPLACE(@a,CHAR(10),' '), ASCII(REPLACE(@a,CHAR(10),' '))

Now please try to explain how your problem is different from what has been suggested.