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From "Michael Neale" <michael.ne...@gmail.com>
Subject Re: Performance problem with references
Date Tue, 12 Sep 2006 09:03:05 GMT
I gather you are referring to saving the node that has all the references as


NodeA ---> NodeX *

Where there are 100 000 NodeX instances. NodeX.getReferences() would then
return just one NodeA (in this case).
So when you save NodeA - you are saving all the references - and I would
expect that to be expensive. However, when you save NodeX - that should be
reasonably fast right?

If that is the case, could you try and "refactor" your repository (is that
the right word ?? ;)  so that there is a child node of NodeA which has all
the references, and if you are making a simple field change to nodeA (not
the references) just save it? (not its child).

Just some thoughts... (I am still learning myself, so this would help me to

On 9/11/06, Christoph Kiehl <kiehl@subshell.com> wrote:
> Hi,
> I'm experiencing an extreme performance/storage problem with references.
> In my
> case I got node(document) that reference another node(state) by reference.
> This
> state node is referenced by about 100.000 document nodes (still growing to
> about
> 1.000.000). As NodeReferences get persisted on every change, there is a
> lot of
> data written to the database on each added reference.
> What is the best way to handle this number of references? Is there a
> better way
> than saving the state uuid as string property an hence give up referencial
> integrity?
> Cheers,
> Christoph

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