On Tue, Jun 18, 2024 at 11:54 AM Dilip Kumar <dilipbal...@gmail.com> wrote: > > On Mon, Jun 17, 2024 at 8:51 PM Robert Haas <robertmh...@gmail.com> wrote: > > > > On Mon, Jun 17, 2024 at 1:42 AM Amit Kapila <amit.kapil...@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. -- With Regards, Amit Kapila.
