SQL mit Vergleich ANzahl Sätzen pro Kunde

[alfredo]

So würde ich da lösen:


with x
as (select gv11fc, count(*) as cnt from gvp011 group by gv11fc),
y as (select gv11fc, count(*) as cnt from gvp011x group by gv11fc)
select x.gv11fc, x.cnt, y.cnt from x
left outer join y on x.gv11fc=y.gv11fc
where x.cnt <> y.cnt

lg