On Tue, Jun 18, 2024 at 12:11 PM Amit Kapila <amit.kapila16@gmail.com> wrote:
>
> On Tue, Jun 18, 2024 at 11:54 AM Dilip Kumar <dilipbalaut@gmail.com> wrote:
> >
> > On Mon, Jun 17, 2024 at 8:51 PM Robert Haas <robertmhaas@gmail.com> wrote:
> > >
> > > On Mon, Jun 17, 2024 at 1:42 AM Amit Kapila <amit.kapila16@gmail.com> wrote:
> > > > The difference w.r.t the existing mechanisms for holding deleted data
> > > > is that we don't know whether we need to hold off the vacuum from
> > > > cleaning up the rows because we can't say with any certainty whether
> > > > other nodes will perform any conflicting operations in the future.
> > > > Using the example we discussed,
> > > > Node A:
> > > > T1: INSERT INTO t (id, value) VALUES (1,1);
> > > > T2: DELETE FROM t WHERE id = 1;
> > > >
> > > > Node B:
> > > > T3: UPDATE t SET value = 2 WHERE id = 1;
> > > >
> > > > Say the order of receiving the commands is T1-T2-T3. We can't predict
> > > > whether we will ever get T-3, so on what basis shall we try to prevent
> > > > vacuum from removing the deleted row?
> > >
> > > The problem arises because T2 and T3 might be applied out of order on
> > > some nodes. Once either one of them has been applied on every node, no
> > > further conflicts are possible.
> >
> > If we decide to skip the update whether the row is missing or deleted,
> > we indeed reach the same end result regardless of the order of T2, T3,
> > and Vacuum. Here's how it looks in each case:
> >
> > Case 1: T1, T2, Vacuum, T3 -> Skip the update for a non-existing row
> > -> end result we do not have a row.
> > Case 2: T1, T2, T3 -> Skip the update for a deleted row -> end result
> > we do not have a row.
> > Case 3: T1, T3, T2 -> deleted the row -> end result we do not have a row.
> >
>
> In case 3, how can deletion be successful? The row required to be
> deleted has already been updated.
Hmm, I was considering this case in the example given by you above[1],
so we have updated some fields of the row with id=1, isn't this row
still detectable by the delete because delete will find this by id=1
as we haven't updated the id? I was making the point w.r.t. the
example used above.
[1]
> > > > Node A:
> > > > T1: INSERT INTO t (id, value) VALUES (1,1);
> > > > T2: DELETE FROM t WHERE id = 1;
> > > >
> > > > Node B:
> > > > T3: UPDATE t SET value = 2 WHERE id = 1;
--
Regards,
Dilip Kumar
EnterpriseDB: http://www.enterprisedb.com
