Re: How to shorten a chain of logically replicated servers

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От Mike Lissner
Тема Re: How to shorten a chain of logically replicated servers
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Msg-id CAMp9=Eyp3YSDsMmzTmUuF3K48h8R=7kb4Jzj-F+m0DYN8WEOiQ@mail.gmail.com
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Ответ на How to shorten a chain of logically replicated servers  (Mike Lissner <mlissner@michaeljaylissner.com>)
Список pgsql-general
Hi, I don't usually like to bump messages on this list, but since I
sent mine on New Year's Eve, I figured I'd better. Anybody have any
ideas about how to accomplish this? I'm pretty stumped (as you can
probably see).

On Tue, Dec 31, 2019 at 3:51 PM Mike Lissner
<mlissner@michaeljaylissner.com> wrote:
>
> Hi, I'm trying to figure out how to shorten a chain of logically
> replicating servers. Right now we have three servers replicating like
> so:
>
> A --> B --> C
>
> And I'd like to remove B from the chain of replication so that I only have:
>
> A --> C
>
> Of course, doing this without losing data is the goal. If the
> replication to C breaks temporarily, that's fine, so long as all the
> changes on A make it to C eventually.
>
> I'm not sure how to proceed with this. My best theory is:
>
> 1. In a transaction, DISABLE the replication from A to B and start a
> new PUBLICATION on A that C will subscribe to in step ③ below. The
> hope is that this will simultaneously stop sending changes to B while
> starting a log of new changes that can later be sent to C.
>
> 2. Let any changes queued on B flush to C. (How to know when they're
> all flushed?)
>
> 3. Subscribe C to the new PUBLICATION created in step ①. Create the
> subscription with copy_data=False. This should send all changes to C
> that hadn't been sent to B, without sending the complete tables.
>
> 4. DROP all replication to/from B (this is just cleanup; the incoming
> changes to B were disabled in step ①, and outgoing changes from B were
> flushed in step ②).
>
> Does this sound even close to the right approach? Logical replication
> can be a bit finicky, so I'd love to have some validation of the
> general approach before I go down this road.
>
> Thanks everybody and happy new year,
>
> Mike



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