getting at namespace data when Parsing Xml on Python 2.6.6 with ElementTree 1.2.6

Question

I have some xml that I need to parse, which is supplied in the following format:

<Pres xmlns:abset="titan:arm:params:xml:ns:keyprov:abset">
<Set>
    <Key>
        <Id>c91e3882-e6f3-41f9-af52-3473a2c4615a</Id>
        <abset:Data>
            <abset:Tag>
                <abset:Value>i need this</abset:Value>
            </abset:Tag>
        </abset:Data>
    </Key>
</Set>
</Pres>

With Python 2.7, I can get at the abset:Value with the following:

xmlstr          = '...'
root            = ElementTree.fromstring(xmlstr)

ns              = {'abset' : 'titan:arm:params:xml:ns:abset'}

keyElement      = root.find("./Set/Key")
value           = keyElement.find("./abset:Data/abset:Tag/abset:Value", ns).text

But in python 2.6, the find command doesn't support the ns argument.

I've tried ValElement = root.find("./Set/Key/abset:Data/abset:Tag/abset:Value") value = ValElement.text

But the error I get is

 keyelement      = root.find("./Set/Key/abset:Data/abset:Tag/abset:Value")
  File "/usr/lib64/python2.6/xml/etree/ElementTree.py", line 330, in find
    return ElementPath.find(self, path)
  File "/usr/lib64/python2.6/xml/etree/ElementPath.py", line 186, in find
    return _compile(path).find(element)
  File "/usr/lib64/python2.6/xml/etree/ElementPath.py", line 176, in _compile
    p = Path(path)
  File "/usr/lib64/python2.6/xml/etree/ElementPath.py", line 93, in __init__
    "expected path separator (%s)" % (op or tag)
SyntaxError: expected path separator (:)

How can I access these elements in python 2.6.6?

Josh Wood · Accepted Answer · 2015-07-22 12:39:33Z

2

You would need to specify the namespace in full, so instead of root.find("./Set/Key/abset:Data/abset:Tag/abset:Value") you should use root.find("./Set/Key/{titan:arm:params:xml:ns}Data/{titan:arm:params:xml:ns}Tag/{titan:arm:params:xml:ns}Value")

answered Jul 22, 2015 at 12:39

Josh Wood

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