Not necessarily faster, but:

select sum(nvl(field1, 1)) as Field1NullsCnt...where isnull(Field1)

or select count(*) ... where isnull(field1)

>Hi everybody,
>
>Suppose, you have a file with 20000 records. Is there a way to check, if field contains NULLs or not for the whole file except for scanning through it and check each field, which would be time consuming...
>
>I just thought about another idea: select sum(iif(isnull(field1,1,0)) as Field1NullsCnt, etc.
> but it would be slow too...
>
>Do you have other suggestions?