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Path of a file stored in Alfresco

camillo

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mikeh

You've got an xpath there - I suspect you're after a display path instead.

Thanks,
Mike

camillo

so, how can i obtain a path in order to parse the xml file?

mikeh

How are you getting the xpath?

Mike

camillo

I call a method getPath which is in APIs written by some friends of mine (that are unreachable now)
So, the question is now: how can i get the path of a resource, e.g a file xml, in order to open it?
Thanks very much for your time!

camillo

In this moment, i can open my xml using this path:
/opt/alfresco/alf_data/contentstore/2009/3/17/11/0/79bd40db-12da-11de-a552-c534a6742d03.bin
where 79bd40db-12da-11de-a552-c534a6742d03.bin is the node representing my xml file.
Now i would like a method or a property that allows me to be free from my filesystem.
So…what's the property or the method i have to call in order to obtain the path for my alf_data directory?
I have a method that returns /2009/3/17/11/0/79bd40db-12da-11de-a552-c534a6742d03.bin, so I need the method that returns /opt/alfresco/alf_data/contentstore
Thanks one more!

jpfi

Hi Camillo,

I must develop a java application able to parse an xml file stored in Alfresco.

What you're doing via your getPath-API call is to find out where the content file in the Alfresco content store is stored…
But, normally you doesn't want to do this. Alfresco provides sth. called repository. This Repo encapsulates a content file (stored in content store) and its metadata (stored in DB).
You can access the content or its metadata using one of the Alfresco APIs.
cheers, Jan

camillo

Hi Jan,

thank for your answer.
In my application i would like know where is the contentstore directory used by alfresco
I know the path of this directory using the repository.properties file, but the problem is I want know this directory using a method called from my java application.
I'm reading Alfresco Repository APIs but I don't find any method or classes that help me.
Can you tell me some classes/methods that could solve my problem?
Thanks

zaizi

You can get the input stream for the XML file by

ContentReader reader = this.contentService.getReader(nodeRef, ContentModel.PROP_CONTENT);‍

where the nodeRef is the NodeRef of the XML file.

camillo

You can get the input stream for the XML file by
ContentReader reader = this.contentService.getReader(nodeRef, ContentModel.PROP_CONTENT);‍
where the nodeRef is the NodeRef of the XML file.

i would link a File object instead of a ContentReader object.
Is this possible?

ebrenes

private File getZipAsFile(String nodeRef) {
    NodeRef node = new NodeRef(nodeRef);
    ContentReader reader = this.contentService.getReader(node, ContentModel.PROP_CONTENT);

    File sourceFile = TempFileProvider.createTempFile(getClass().getSimpleName() + "_source_", ".zip");
    reader.getContent(sourceFile);

    return sourceFile;

}

You can change the extension file to whatever you need.