This statement select * from A full outer join B on nvl(a.id,0)=nvl(b.id,0)
generates a wrong graph (hedge is lost)
and this one: **select * from A full outer join (select * from T) B on a.id=b.id **
doesn't generate any graph at all
Regards PAscal
Hi,
Problem is with simplifier . Both theese work:
**select * from A full outer join B on a.id=b.id select * from A full outer join B on 0 and g(a.id) = f(b.id,0) **
-Alex
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This statement
select * from A full outer join B
on nvl(a.id,0)=nvl(b.id,0)
generates a wrong graph (hedge is lost)
and this one:
**select * from A full outer join (select * from T) B
on a.id=b.id **
doesn't generate any graph at all
Regards
PAscal
Hi,
Problem is with simplifier . Both theese work:
**select * from A full outer join B on a.id=b.id
select * from A full outer join B on 0 and g(a.id) = f(b.id,0) **
-Alex