How to get an xpathContext from an xmlNode in python

Question

big fan of xpath on .net, and sax in python, but first time using xpath in python.

I have a small script, that uses xpath to select some nodes from a doc, iterates through them, and then ideally uses xpath again to get the relevant data from them. However I can't get that last bit, once I have the xmlNode I cannot get a context from it.

import libxml2
import urllib

doc = libxml2.parseDoc(
        urllib.urlopen('http://somemagicwebservice.com/').read())
ctxt = doc.xpathNewContext()
listitems = ctxt.xpathEval('//List/ListItem')
for item in listitems:
    itemctxt = item.xpathNewContext()
    title = itemctxt.xpathEval('//ItemAttributes/Title')
    asin = itemctxt.xpathEval('//Item/ASIN')
    itemctxc.xpathFreeContext()
ctxt.xpathFreeContext()
doc.freeDoc()

However the itemctxt = item.xpathNewContext() bit fails with

itemctxt = item.xpathNewContext()
AttributeError: xmlNode instance has no attribute 'xpathNewContext'

Any ideas how to use xpath on a xmlNode? I can't find any good online info. Thanks

Lennart Regebro · Accepted Answer · 2009-07-05 14:06:19Z

2

I don't think an XPathContext makes sense on an element? Try creating a new XPathContext, and setting it's node to the current element.

That said, I haven't used libxml2 directly, so it's a bit of a wild guess. I typically uses lxml, that exposes an ElementTree API around libxml2 and libxslt. It's much easier to use, and does indeed allow xpath() on elements. Of course, if you already have a lot of code using libxml2 you probably don't want to switch, but in that case you might want to take a look at lxmls source to see how it does it.

http://codespeak.net/svn/lxml/trunk/src/lxml/xpath.pxi

http://codespeak.net/svn/lxml/trunk/src/lxml/_elementpath.py

Seems good starting places.

answered Jul 5, 2009 at 14:06

Lennart Regebro

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2 Comments

Kurt Over a year ago

Well I had similar problems with lxml, but it turns out the problem was actually related to how xpath deals with namespaces. lxml's ElementPath simplified things a bit. Thanks.

Kurt Over a year ago

If anyone wants to see what I ended up with, check out bitbucket.org/kurthaeusler/getwishlist/src/tip/getwishlist.py

Community · Accepted Answer · 2017-05-23 12:06:41Z

2

https://stackoverflow.com/a/3379708/288875 suggests to call setContextNode(..) on a newly created context:

itemctxt = doc.xpathNewContext()

for item in listitems:
    itemctxt.setContextNode(item)
    title = itemctxt.xpathEval('.//ItemAttributes/Title')

    ...

itemctxt.xpathFreeContext()

In the version of python libxml (2.9.1) which I'm currently using it turns out that one can even call:

item.xpathEval('.//ItemAttributes/Title')

Note that you'll have to add a dot at the beginning of the xpath expressions .// (instead of //), otherwise you'll get search results relative to the document root.

edited May 23, 2017 at 12:06

CommunityBot

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answered Sep 17, 2015 at 11:20

Andre Holzner

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