Am 04.11.2010 20:53, schrieb A B:
> First a more general question: is there any clever way to do two
> selects A and B and then return the result
> A union ( B \ (A intersect B)) ( \ is "set subtraction")
> Any ideas besides writing the explicit queries? I guess one has to
> lock the table to get the same result on both selects unless one can
> cache the result of A and B so you don't run it twice?
> Then the actual problem at hand, which with extra details might result
> in another solution then the one from above.
> A looks like select a.id,a.name,true from X
> B looks like select b.id,b.name,false from Y
( A intersect B ) is allways the empty set because of the boolean
column, isn't it?
Even if there were tupels (a.id, a.name) = (b.id, b.name)
still (a.id, a.name, TRUE) <> (b.id, b.name, FALSE)
because of TRUE <> FALSE
So ( A intersect B ) = ()
Therefore ( B \ () ) = B
So you end up with A union B as Tom said.
The clever way to do the stuff w/o to many A- and B-subselects may be to
create two temporary views and use those. There might be a performance
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