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No simple path to NodeRef?

Question asked by adepue on Jul 21, 2005
After looking through the NodeService interface, I'm guessing there is no simple method that can take a simple path as input and produce the matching NodeRef?  I know there is the XPath based methods, but this isn't "simple" due to my usage scenario.  I'm reimplementing a simple "DocumentRepositoryDAO" that takes simple paths as input (simple, as in "someFolder/possibleOtherFolders/file").  I can't just pass these straight to XPath, as they may contain characters that XPath doesn't like (or worse, provide a door for a crafty user to inject unfriendly XPath into the query).  So that leaves me with manually parsing the path and walking the hierarchy, or manually converting the path to XPath.  Are there any plans to introduce some utility method to easily get the NodeRef for a simple path?  Or maybe such a method exists and I just haven't found it?

Thanks,
  Andy

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