>Hi Gregory,
>
>(X.X)
>
>Ich denke, jetzt brauche ich einen Schnaps.


Have two !

>Works perfect, But I need to think through this
>
>>

>>	select c3 ;
>>		from Schnaps ;
>>		group by 1 ;
>>		having ( count( distinct c1 + c2 ) > 1) ;
>>

>

Count(*) is not well documented in the help

This may help http://msdn.microsoft.com/en-us/library/ms175997.aspx

	create cursor pp ;
	(	c1 c(1) null ;
	)
	
	insert into pp values( null)
	
	insert into pp values( 'a')
	insert into pp values( 'b')
	insert into pp values( 'b')

	
	
	select count( c1 ) from pp && 3 non null
	select count( all c1 ) from pp && 3 non null
	
	select count( distinct  c1 ) from pp && 2 non null