How to find 'x' number of days from 'yyyymmdd' format?

[DBFan]

I need to find out the count of number of records older than 100 days from a table having 'order_date' as yyyymmdd format eg. 20041115. Thanks in advance.

[MAK]

select datediff(dd,'20031111','20041201')

[MAK]

select datediff(dd,'20031111','20041201')
select datediff(dd,'2003/11/11','2004/12/01')
select datediff(dd,'2003-11-11 12:00:00','2004-12-01 19:25:00')
select datediff(dd,'2003-11-11 23:00:00','2004-12-01 19:25:00')

[DBFan]

I will not be knowing the dates so as to hard code them in the query..instead I need to find out how many records are there which are older than say 100days from a table like this:

order_number order_date
1001 20041023
1002 20041023
10003 20041023
7687 20041023
7909 20041023
7699 20041023
9878 20041023

Any help?

[Rawhide]

Select Count(*)
From myTable
Where DateDiff(dd, order_date, getdate()) > 100

[DBFan]

I am getting this error as order_date is declared as int for some reason:

Server: Msg 8115, Level 16, State 2, Line 1
Arithmetic overflow error converting expression to data type datetime.

[Rawhide]

Set DateFormat ymd

Select Count(*)
From myTable
Where DateDiff(dd, Cast(order_date as varchar)), getdate()) > 100

[DBFan]

Getting the following error again:

Server: Msg 174, Level 15, State 1, Line 3
The datediff function requires 3 arguments.

[Rawhide]

Sorry, had a typo in my code:

Select Count(*)
From myTable
Where DateDiff(dd, Cast(order_date as varchar), getdate()) > 100

[DBFan]

It worked! Thank you Rawhide and everyone :-)