Обсуждение: How to do A union (B - ( A intersect B )) or sort of :-)

Hi.

Sorry for the subject, but I knew no words to describe it better.
I will try to explain

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

can one then use the differences in boolean values in some clever way,
like  OR  operation?

A B <gentosaker@gmail.com> writes:
> Sorry for the subject, but I knew no words to describe it better.
> I will try to explain

> 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")

Ummm ... isn't that just a UNION?  Anyway, SQL's features for this
sort of thing are described here:
http://www.postgresql.org/docs/9.0/static/queries-union.html

> 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?

Both arms of a UNION or similar query would be reading the table with
the same snapshot, so there wouldn't be any need for a lock to get
consistent results.  You may also need to spend some time reading
http://www.postgresql.org/docs/9.0/static/mvcc.html

            regards, tom lane

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
loss though.