Re: lateral join with union all - Mailing list pgsql-general

From David G. Johnston
Subject Re: lateral join with union all
Date
Msg-id CAKFQuwa+vX7CbHGCJF9anj4PPveQQj47S=pjJodZXraHKZ7e=g@mail.gmail.com
Whole thread Raw
In response to lateral join with union all  (jian he <jian.universality@gmail.com>)
List pgsql-general
On Mon, Aug 15, 2022 at 8:53 AM jian he <jian.universality@gmail.com> wrote:
select * from
(
    (select 2 as v )
    union all
    (select 3 as v)
) as q1
cross join lateral
(   (select * from
    ((select 4 as v) union all
    (select 5 as v)) as q3
    )
    union all
    (select q1.v)
) as  q2;

I thought q1 will be materialized as a constant set and will be equivalent as  select 2 union all select 3;
Then It will  have 8 (2 * 4) rows total. Then It will be like {2,3} cross join with {2,3,4,5}

But Here the actual result(return 6 rows) feels like two separate queries(A,B) then union together.
QueryA: (select 2 as v ) cross join lateral (.....)
QueryB: (select 3 as v ) cross join lateral (.....)
Query A 3 row + Query B 3 row. So the total is 6 rows.

Then I feel a little bit confused.


Lateral is literally a FOR EACH row construct.  So q2 is evaluated for the first row in q1, then it is evaluated for the second row of q1.  Which produces 6 rows (4 from q2 literal rows plus two more by copying the current row of q1 into a new row within q2 - twice).

CROSS JOIN here is a mis-nomer, and I personally avoid using it for that reason.  You are really doing an inner join between a single row from q1 and each and every row produced by evaluating q2 in the context of that q1 row (it's a bit easier to understand if you have a function lateral as opposed to a subquery, but the effects are identical).

q1 INNER JOIN LATERAL AS q2 ON true

Where q2 can use the columns of q1 in producing its output.

David J.

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