[XML-SIG] get the abolute path for a node

[Alexandre CONRAD]

> Does this help?
> 
> def abs_path( node ):
>     successors = 1
>     parent = node.previousSibling
>     while parent:
>         if parent.nodeName == node.nodeName: successors += 1
>         parent = parent.previousSibling
>     name = node.nodeName == '#text' and 'text()' or node.nodeName
>     path = successors>1 and '/%s[%s]'%(name,successors) or '/%s'%name
>     if node.parentNode and node.parentNode.nodeName != '#document':
>         return abs_path( node.parentNode )+path
>     return path


Because I always strip out spaces in XML documents, and because I want 
to show the 1st node with node[1], I changed your code so:

- name = node.nodeName == '#text' and 'text()' or node.nodeName
- path = successors>1 and '/%s[%s]'%(name,successors) or '/%s'%name
+ path = '/%s[%s]' % (node.nodeName, successors)


This is function is pretty neat. But still, there is 1 more little thing 
that I'm having a hard time figuring out how to fix. I keep getting 
"/playlist[2]" as the root node. I can't have 2 root nodes anyway...

<?xml version='1.0' encoding='UTF-8'?>
<playlist>
   <group> <-- shows: /playlist[2]/group[1]
     <video>foo.mpg</video> <-- shows: /playlist[2]/group[1]/video[1]
     <video>bar.mpg</video> <-- shows: /playlist[2]/group[1]/video[2]
   </group>
</playlist>

And this looping code inside the function everytime makes me loose track 
of what's doing on. Well done though.

Best regards,
-- 
Alexandre CONRAD - TLV
Research & Development
tel : +33 1 30 80 55 05
fax : +33 1 30 56 55 06
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78860 - SAINT NOM LA BRETECHE
FRANCE