On Wed, Sep 22, 2021 at 1:50 PM Amit Kapila <amit.kapila16@gmail.com> wrote:
>
> On Wed, Sep 22, 2021 at 6:42 AM Ajin Cherian <itsajin@gmail.com> wrote:
> >
>
> Why do you think that the second assumption (if there is an old tuple
> it will contain all RI key fields.) is broken? It seems to me even
> when we are planning to include unchanged toast as part of old_key, it
> will contain all the key columns, isn't that true?
Yes, I assumed wrongly. Just checked. What you say is correct.
>
> > I think we
> > still need to deform both old tuple and new tuple, just to handle this case.
> >
>
> Yeah, but we will anyway talking about saving that cost for later if
> we decide to send that tuple. I think we can further try to optimize
> it by first checking whether the new tuple has any toasted value, if
> so then only we need this extra pass of deforming.
Ok, I will go ahead with this approach.
>
> > There is currently logic in ReorderBufferToastReplace() which already
> > deforms the new tuple
> > to detoast changed toasted fields in the new tuple. I think if we can
> > enhance this logic for our
> > purpose, then we can avoid an extra deform of the new tuple.
> > But I think you had earlier indicated that having untoasted unchanged
> > values in the new tuple
> > can be bothersome.
> >
>
> I think it will be too costly on the subscriber side during apply
> because it will update all the unchanged toasted values which will
> lead to extra writes both for WAL and data.
>
Ok, agreed.
regards,
Ajin Cherian
Fujitsu Australia