From: | Laurenz Albe <laurenz(dot)albe(at)cybertec(dot)at> |
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To: | Bharath Rupireddy <bharath(dot)rupireddyforpostgres(at)gmail(dot)com>, PostgreSQL Hackers <pgsql-hackers(at)lists(dot)postgresql(dot)org> |
Cc: | SATYANARAYANA NARLAPURAM <satyanarlapuram(at)gmail(dot)com> |
Subject: | Re: An attempt to avoid locally-committed-but-not-replicated-to-standby-transactions in synchronous replication |
Date: | 2022-04-26 06:26:59 |
Message-ID: | 9290b55b6ae2b04e002ca9dadadd1cca09461482.camel@cybertec.at |
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Lists: | pgsql-hackers |
On Mon, 2022-04-25 at 19:51 +0530, Bharath Rupireddy wrote:
> With synchronous replication typically all the transactions (txns)
> first locally get committed, then streamed to the sync standbys and
> the backend that generated the transaction will wait for ack from sync
> standbys. While waiting for ack, it may happen that the query or the
> txn gets canceled (QueryCancelPending is true) or the waiting backend
> is asked to exit (ProcDiePending is true). In either of these cases,
> the wait for ack gets canceled and leaves the txn in an inconsistent
> state [...]
>
> Here's a proposal (mentioned previously by Satya [1]) to avoid the
> above problems:
> 1) Wait a configurable amount of time before canceling the sync
> replication by the backends i.e. delay processing of
> QueryCancelPending and ProcDiePending in Introduced a new timeout GUC
> synchronous_replication_naptime_before_cancel, when set, it will let
> the backends wait for the ack before canceling the synchronous
> replication so that the transaction can be available in sync standbys
> as well.
> 2) Wait for sync standbys to catch up upon restart after the crash or
> in the next txn after the old locally committed txn was canceled.
While this may mitigate the problem, I don't think it will deal with
all the cases which could cause a transaction to end up committed locally,
but not on the synchronous standby. I think that only using the full
power of two-phase commit can make this bulletproof.
Is it worth adding additional complexity that is not a complete solution?
Yours,
Laurenz Albe
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