CODEWITHSUNDEEP
pythonxmllxmlxml-namespaces
Laurent LAPORTE
Laurent LAPORTE

Reputation: 22952

How to insert an attribute with the right namespace prefix using lxml

Is there a way, using lxml, to insert XML attributes with the right namespace?

For instance, I want to use XLink to insert links in a XML document. All I need to do is to insert {http://www.w3.org/1999/xlink}href attributes in some elements. I would like to use xlink prefix, but lxml generates prefixes like "ns0", "ns1"…

Here is what I tried:

from lxml import etree

#: Name (and namespace) of the *href* attribute use to insert links.
HREF_ATTR = etree.QName("http://www.w3.org/1999/xlink", "href").text

content = """\
<body>
<p>Link to <span>StackOverflow</span></p>
<p>Link to <span>Google</span></p>
</body>
"""

targets = ["https://stackoverflow.com", "https://www.google.fr"]
body_elem = etree.XML(content)
for span_elem, target in zip(body_elem.iter("span"), targets):
    span_elem.attrib[HREF_ATTR] = target

etree.dump(body_elem)

The dump looks like this:

<body>
<p>link to <span xmlns:ns0="http://www.w3.org/1999/xlink"
                 ns0:href="https://stackoverflow.com">stackoverflow</span></p>
<p>link to <span xmlns:ns1="http://www.w3.org/1999/xlink"
                 ns1:href="https://www.google.fr">google</span></p>
</body>

I found a way to factorize the namespaces by inserting and deleting an attribute in the root element, like this:

# trick to declare the XLink namespace globally (only one time).
body_elem = etree.XML(content)
body_elem.attrib[HREF_ATTR] = ""
del body_elem.attrib[HREF_ATTR]

targets = ["https://stackoverflow.com", "https://www.google.fr"]
for span_elem, target in zip(body_elem.iter("span"), targets):
    span_elem.attrib[HREF_ATTR] = target

etree.dump(body_elem)

It's ugly, but it works and I only need to do it one time. I get:

<body xmlns:ns0="http://www.w3.org/1999/xlink">
<p>Link to <span ns0:href="https://stackoverflow.com">StackOverflow</span></p>
<p>Link to <span ns0:href="https://www.google.fr">Google</span></p>
</body>

But the problem remains: how can I turn this "ns0" prefix into "xlink"?

Upvotes: 1

Views: 698

Answers (1)

Laurent LAPORTE
Laurent LAPORTE

Reputation: 22952

Using register_namespace as suggested by @mzjn:

etree.register_namespace("xlink", "http://www.w3.org/1999/xlink")

# trick to declare the XLink namespace globally (only one time).
body_elem = etree.XML(content)
body_elem.attrib[HREF_ATTR] = ""
del body_elem.attrib[HREF_ATTR]

targets = ["https://stackoverflow.com", "https://www.google.fr"]
for span_elem, target in zip(body_elem.iter("span"), targets):
    span_elem.attrib[HREF_ATTR] = target

etree.dump(body_elem)

The result is what I expected:

<body xmlns:xlink="http://www.w3.org/1999/xlink">
<p>Link to <span xlink:href="https://stackoverflow.com">StackOverflow</span></p>
<p>Link to <span xlink:href="https://www.google.fr">Google</span></p>
</body>

Upvotes: 2

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