namespace error lxml xpath python

Question

I am transforming word documents to xml to compare them using the following code:

word = win32com.client.Dispatch('Word.Application')
wd = word.Documents.Open(inFile)
# Converts the word infile to xml outfile
wd.SaveAs(outFile,11)
wd.Close()
dom=parse(outFile)

The xml file I get looks like:

<?xml version="1.0" encoding="utf-8"?>
<?mso-application progid="Word.Document"?>
<w:wordDocument w:embeddedObjPresent="no" w:macrosPresent="no" w:ocxPresent="no" xml:space="preserve" xmlns:aml="http://schemas.microsoft.com/aml/2001/core" xmlns:dt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882" xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:sl="http://schemas.microsoft.com/schemaLibrary/2003/core" xmlns:v="urn:schemas-microsoft-com:vml" xmlns:w="http://schemas.microsoft.com/office/word/2003/wordml" xmlns:w10="urn:schemas-microsoft-com:office:word" xmlns:wne="http://schemas.microsoft.com/office/word/2006/wordml" xmlns:wpc="http://schemas.microsoft.com/office/word/2010/wordprocessingCanvas" xmlns:wsp="http://schemas.microsoft.com/office/word/2003/wordml/sp2" xmlns:wx="http://schemas.microsoft.com/office/word/2003/auxHint">
    <w:ignoreSubtree w:val="http://schemas.microsoft.com/office/word/2003/wordml/sp2"/>
    <w:shapeDefaults>
        <o:shapedefaults spidmax="1027" v:ext="edit"/>
        <o:shapelayout v:ext="edit">
            <o:idmap data="1" v:ext="edit"/>
        </o:shapelayout>
    </w:shapeDefaults>
    <w:body>
        <wx:sect>
            <w:tbl>

            <w:tblGrid>
                <w:gridCol w:w="200"/>
                                       ...
            </w:tblGrid>

                <w:pict>
                        <v:shapetype coordsize="21600,21600" filled="f" id="_x0000_t75" o:preferrelative="t" o:spt="75" path="m@4@5l@4@11@9@11@9@5xe" stroked="f">
                            <v:stroke joinstyle="miter"/>
                            <v:formulas>
                                <v:f eqn="if lineDrawn pixelLineWidth 0"/>
                                ...
                            </v:formulas>
                            <v:path gradientshapeok="t" o:connecttype="rect" o:extrusionok="f"/>
                            <o:lock aspectratio="t" v:ext="edit"/>
                        </v:shapetype>
                        <v:shape id="Picture" o:spid="_x0000_s1026" style="position:absolute;left:0;text-align:left;margin-left:0;margin-top:0;width:400pt;height:40pt;z-index:1;visibility:visible;mso-wrap-style:square;mso-wrap-distance-left:0;mso-wrap-distance-top:0;mso-wrap-distance-right:0;mso-wrap-distance-bottom:0;mso-position-horizontal:left;mso-position-horizontal-relative:text;mso-position-vertical:absolute;mso-position-vertical-relative:line" type="#_x0000_t75">
                            <v:imagedata o:title="" src="wordml://03000001.png"/>
                            <w10:wrap anchory="line"/>
                            <w10:anchorlock/>
                        </v:shape>
                </w:pict> 
                                      ...

I can't use xpath function (lxml library) when I try for example :

import lxml.etree as et
tree = et.parse(xmlFile)
for elt in tree.xpath("//w:gridCol"):
     elt.getparent().remove(elt)

I get the following error:

 for elt in tree.xpath("//w:gridCol"):
  File "lxml.etree.pyx", line 2029, in lxml.etree._ElementTree.xpath (src/lxml/lxml.etree.c:45934)
  File "xpath.pxi", line 379, in lxml.etree.XPathDocumentEvaluator.__call__ (src/lxml/lxml.etree.c:114389)
  File "xpath.pxi", line 242, in lxml.etree._XPathEvaluatorBase._handle_result (src/lxml/lxml.etree.c:113063)
  File "xpath.pxi", line 227, in lxml.etree._XPathEvaluatorBase._raise_eval_error (src/lxml/lxml.etree.c:112894)
XPathEvalError: Undefined namespace prefix

I did some research and I guess it's a namespace matter, but I don't know how to fix it?

Register the namespace (see http://stackoverflow.com/a/14853417/3190413) or ignore them using `//*[local-name()='gridCol']` — helderdarocha, Jun 04 '14 at 13:39
@helderdarocha ignoring namespace works fine Thanks for the answer — user3328690, Jun 04 '14 at 16:30

score 12 · Answer 1 · edited Aug 09 '17 at 08:39

12

In this code:

for elt in tree.xpath("//w:gridCol"):

w: isn't a namespace; it's a namespace prefix which is effectively shorthand for the actual namespace, http://schemas.microsoft.com/office/word/2003/wordml. If you want to search for elements in this namespace using the xpath method, you need to provide it with a mapping of namespace prefixes to namespaces:

tree.xpath("//w:gridCol", namespaces={
  'w': 'http://schemas.microsoft.com/office/word/2003/wordml',
  })

Also, note that there is no requirement that you use the same namespace prefix. The following would find the same elements:

tree.xpath("//bob:gridCol", namespaces={
  'bob': 'http://schemas.microsoft.com/office/word/2003/wordml'
  })

edited Aug 09 '17 at 08:39

Alfe

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answered May 08 '15 at 12:15

larsks

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2

The `elt` already knows the namespaces and their mapping (see `elt.nsmap`), so why doesn't it use them? – Alfe Aug 08 '17 at 14:17
That seems like a good question for the developer. I haven't examined the code in that level of detail. – larsks Aug 08 '17 at 21:18
2

I reconsidered this with a coworker and we came to the conclusion that it wouldn't be correct to reuse the `.nsmap`. The ns-prefixes can change from instance to instance since the sender of the XML can choose them freely. If we write code using a fixed ns-prefix, that's okay as long as we also provide a fixed map which associates this ns-prefix with the namespace-uri. Relying on the `.nsmap` would just be insufficient. – Alfe Aug 08 '17 at 23:41

namespace error lxml xpath python

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