Horton

Re: when the startup process doesn't (logging startup delays)

Bharath Rupireddy <bharath.rupireddyforpostgres@gmail.com>

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From: Bharath Rupireddy <bharath.rupireddyforpostgres@gmail.com>
To: Robert Haas <robertmhaas@gmail.com>
Cc: Simon Riggs <simon.riggs@enterprisedb.com>, Thomas Munro <thomas.munro@gmail.com>, Nitin Jadhav <nitinjadhavpostgres@gmail.com>, Tom Lane <tgl@sss.pgh.pa.us>, Alvaro Herrera <alvherre@alvh.no-ip.org>, Justin Pryzby <pryzby@telsasoft.com>, Michael Paquier <michael@paquier.xyz>, Andres Freund <andres@anarazel.de>, Stephen Frost <sfrost@snowman.net>, Magnus Hagander <magnus@hagander.net>, "pgsql-hackers@postgresql.org" <pgsql-hackers@postgresql.org>
Date: 2022-11-16T06:47:13Z
Lists: pgsql-hackers

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On Tue, Nov 15, 2022 at 10:55 PM Robert Haas <robertmhaas@gmail.com> wrote:
>
> On Tue, Nov 15, 2022 at 8:33 AM Bharath Rupireddy
> <bharath.rupireddyforpostgres@gmail.com> wrote:
> > Please review the v2 patch.
>
> It seems to me that this will call disable_startup_progress_timeout
> once per WAL record, which seems like an unnecessary expense. How
> about leaving the code inside the loop just as we have it, and putting
> if (StandbyMode) disable_startup_progress_timeout() before entering
> the loop?

That can be done, only if we can disable the timeout in another place
when the StandbyMode is set to true in ReadRecord(), that is, after
the standby server finishes crash recovery and enters standby mode.

I'm attaching the v3 patch for further review. Please find the CF
entry here - https://commitfest.postgresql.org/41/4012/.

--
Bharath Rupireddy
PostgreSQL Contributors Team
RDS Open Source Databases
Amazon Web Services: https://aws.amazon.com

Commits

  1. Un-revert "Disable STARTUP_PROGRESS_TIMEOUT in standby mode."

  2. Revert "Disable STARTUP_PROGRESS_TIMEOUT in standby mode."

  3. Disable STARTUP_PROGRESS_TIMEOUT in standby mode.

  4. Fix race condition in startup progress reporting.

  5. Report progress of startup operations that take a long time.

  6. Add enable_timeout_every() to fire the same timeout repeatedly.