From: | Stephan Szabo <sszabo(at)megazone23(dot)bigpanda(dot)com> |
---|---|
To: | Rod Taylor <rbt(at)rbt(dot)ca> |
Cc: | PostgreSQL Development <pgsql-hackers(at)postgresql(dot)org> |
Subject: | Re: Foreign key quandries |
Date: | 2003-03-01 07:38:23 |
Message-ID: | 20030228232939.X10095-100000@megazone23.bigpanda.com |
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Thread: | |
Lists: | pgsql-hackers |
On 1 Mar 2003, Rod Taylor wrote:
> Gah, hit wrong key combination and the email sent early.
>
> Anyway, after that 'sleep' mess at the bottom is:
> T1 or T2: Sleeping too long -- lets run deadlock detection code
> T1 or T2: Kill off random participant of deadlock.
>
> The other participant is then allowed to continue their work.
>
> > Isn't the differentiation going to happen automatically?
The problem is that in case 2, both tuples 2 and 3 are already removed
before either foreign key check runs, so when T1 adds the value 3
row and checks the pk table it will find that its pk row has been
modified. If the ordering went, delete 2 - check 2, delete 3 - check
3, this wouldn't be a problem, but then that'd fail in a
spec-non-compliant way if row 2 refered to row 3.
> > In case 2:
> >
> > T1: create fk tuple (uncommitted) -> value 2
* T1: scan through pk table, no problems
> > T2: delete pk tuple value 2
* T2: delete pk tuple value 3
> > T2: scan through fk table, find uncommitted tuple value 2 ... sleep
> > T2: scan through fk table, find uncommitted tuple value 2 ... sleep
> > T2: scan through fk table, find uncommitted tuple value 2 ... sleep
> > T1: create fk tuple (uncommitted) -> value 3
* T1: scan through pk table, find modified tuple value 3 ...
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