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From "Jonathan Ellis (JIRA)" <j...@apache.org>
Subject [jira] [Updated] (CASSANDRA-2610) Have the repair of a range repair *all* the replica for that range
Date Thu, 01 Sep 2011 16:18:09 GMT

     [ https://issues.apache.org/jira/browse/CASSANDRA-2610?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel
]

Jonathan Ellis updated CASSANDRA-2610:
--------------------------------------

    Attachment: 0003-cleanup-and-fix-private-reference.patch

03 fixes a reference to private UUIDGen.instance field, and cleans up RepairJob to (1) use
concurrent structures instead of synchronized and (2) track Differencer objects directly instead
of using Pair<InetAddress, InetAddress> as proxies.  Also adds some comments.

otherwise, lgtm.

> Have the repair of a range repair *all* the replica for that range
> ------------------------------------------------------------------
>
>                 Key: CASSANDRA-2610
>                 URL: https://issues.apache.org/jira/browse/CASSANDRA-2610
>             Project: Cassandra
>          Issue Type: Improvement
>          Components: Core
>    Affects Versions: 0.8 beta 1
>            Reporter: Sylvain Lebresne
>            Assignee: Sylvain Lebresne
>            Priority: Minor
>             Fix For: 1.0
>
>         Attachments: 0001-Make-repair-repair-all-hosts-v2.patch, 0001-Make-repair-repair-all-hosts.patch,
0002-Cleanup-log-messages-v2.patch, 0003-cleanup-and-fix-private-reference.patch
>
>   Original Estimate: 8h
>  Remaining Estimate: 8h
>
> Say you have a range R whose replica for that range are A, B and C. If you run repair
on node A for that range R, when the repair end you only know that A is fully repaired. B
and C are not. That is B and C are up to date with A before the repair, but are not up to
date with one another.
> It makes it a pain to schedule "optimal" cluster repairs, that is repairing a full cluster
without doing work twice (because you would have still have to run a repair on B or C, which
will make A, B and C redo a validation compaction on R, and with more replica it's even more
annoying).
> However it is fairly easy during the first repair on A to have him compare all the merkle
trees, i.e the ones for B and C, and ask to B or C to stream between them whichever the differences
they have. 

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