Sorry. Accidental send my previous response.
Use node walking, starting from rootNode (session.getRootNode()).
Or if you have defined your node structure with a certain pattern, then
you can follow the pattern to reach your node quickly. For example, you
map a file in a local file system to a JCR repository node with a
pattern:
File system side: folder1/filder2/file1
JCR Node side: get appRootNode of your application, then folder1Node =
appRootNode.getNode("folder1");
folder2Node = folder1Node.getNode("folder2");
file1Node = folder2Node.getNode("file1");
Or
file1Node = appRootNode.getNode("folder1/folder2/file1"); -- like Alex
suggestion
Usually you have an index that persist somewhere (database or file
system) to refer to your JCR node, so you can do this kind of quick
mapping.
-----Original Message-----
From: Akil Ali [mailto:[email protected]]
Sent: Thursday, February 26, 2009 10:57 AM
To: [email protected]
Subject: Re: How to retrieve a node from a hierarchy?
Sorry to say, i dont want to use neither XPATH nor UUID. so what would
be the other solution to get the node.
Akil
Alexander Klimetschek wrote:
>
> On Thu, Feb 26, 2009 at 4:42 PM, Akil Ali <[email protected]>
> wrote:
>> Sorry sir i dont have the information that NODE D is under A/B/C. so
>> how to get the NODE object of NODE D. without prior knowledge that D
>> is under A/B/C.
>
> Well, you could either navigate to it by looking for whatever marks
> that node as "D" (properties, node type, node name) or do the same
> with an xpath query. Typically it has a node type, let's say
> "my:dtype", so the query would be:
>
> //element(*, my:dtype)
>
> This gives you all nodes of type "my:dtype". If you have some
> additional property you want to check for, you can do this:
>
> //element(*, my:dtype)[...@myprop='foobar']
>
> which gives you all my:dtype nodes with a property set to the value
> "foobar".
>
> Regards,
> Alex
>
> --
> Alexander Klimetschek
> [email protected]
>
>
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