Denis Perchine <dyp(at)perchine(dot)com> writes:
>> Now that I think about it, this code could do a two-key scan backwards
>> and stop after finding the first (last) valid tuple, but that's more
>> than a one-line change.
> Actual logic is to find the maximum of pageno, for specified oid.
> I do index scan on 2-keys index, specifying only one key as constraint...
> If I do index scan forward I will get the smallest pageno first...
> Otherwise I get the highest pageno... And this is what I want...
Hmm ... probably right, but the loop logic doesn't behave that way
right now.
regards, tom lane