Delete hyperlinks from word document

I am wanting to delete a hyperlink from docx. The xml looks similar to:

<w:p w:rsidP="00893654" w:rsidRDefault="00A04494" w:rsidRPr="00893654" w:rsidR="002B1D7E">
  <w:r w:rsidRPr="00893654">
    <w:t xml:space="preserve">Here is a text section. Hyperlink here: </w:t>
  </w:r>
  <w:hyperlink r:id="rId12" w:history="1" w:tgtFrame="_blank">
    <w:r w:rsidRPr="00893654">
       <w:t xml:space="preserve">This is the Hyperlink.  </w:t>
     </w:r> 
  </w:hyperlink>
</w:p>

I see it in the paragraph._element.xml, but it isn’t editable. I’ve been able to add xml tags in runs using the docx.oxml library but I haven’t seen a way to edit existing xml tags. Worst case would be to unzip the word file, and edit it via lxml and then rezip it. But I would rather have a more elegant solution. Ideas??

[scanny]

I’d say the job is this:

• Promote the enclosed w:r element to a peer of w:hyperlink
• Remove the relationship "rId12" from the document part (or other story part like header).
• Remove the w:hyperlink element.

You probably need to start by getting references to the various elements, probably starting with the w:p container. This is simple enough once you’ve identified the paragraph:

p = paragraph._p

The w:hyperlink is a child, so something like:

hyperlink = p.xpath("./w:hyperlink")[0]

If you don’t care about the enclosed run (the hyperlink “label”) you can delete it with the hyperlink element:

p.remove(hyperlink)

If you want to preserve it, you’ll need to promote it (like move it out of being enclosed by hyperlink):

label_r = p.xpath("./w:hyperlink/w:r")[0]
hyperlink.add_previous(label_r)
p.remove(hyperlink)

The last bit is the relationship, which might be okay to leave dangling, worth a try, or maybe you have to remove it:

from docx.oxml.ns import qn  # ---for "qualified name", aka. Clark-notation tagname---
hyperlink_rel_id = hyperlink.get(qn("r:id"))
document_part = document.part
docuemnt_part.drop_rel(hyperlink_rel_id)

Of course all this uses internals so comes with some risk of brittle breakage with new releases, but these parts have been stable for some time so probably work fine indefinitely.

[yadneshSalvi]

I’d say the job is this:

• Promote the enclosed w:r element to a peer of w:hyperlink
• Remove the relationship "rId12" from the document part (or other story part like header).
• Remove the w:hyperlink element.

You probably need to start by getting references to the various elements, probably starting with the w:p container. This is simple enough once you’ve identified the paragraph:

p = paragraph._p

The w:hyperlink is a child, so something like:

hyperlink = p.xpath("./w:hyperlink")[0]

If you don’t care about the enclosed run (the hyperlink “label”) you can delete it with the hyperlink element:

p.remove(hyperlink)

If you want to preserve it, you’ll need to promote it (like move it out of being enclosed by hyperlink):

label_r = p.xpath("./w:hyperlink/w:r")[0]
hyperlink.add_previous(label_r)
p.remove(hyperlink)

The last bit is the relationship, which might be okay to leave dangling, worth a try, or maybe you have to remove it:

from docx.oxml.ns import qn  # ---for "qualified name", aka. Clark-notation tagname---
hyperlink_rel_id = hyperlink.get(qn("r:id"))
document_part = document.part
docuemnt_part.drop_rel(hyperlink_rel_id)

Of course all this uses internals so comes with some risk of brittle breakage with new releases, but these parts have been stable for some time so probably work fine indefinitely.

This worked well, just had to use ‘addprevious’ in place of ‘add_previous’ wrote in this comment because I spend an hour finding out why it was failing.

Also to replace multiple hyperlinks in a paragraph with text you can try the following for loop

if paragraph._p.xpath("./w:hyperlink"):
    for hyperlink, label_r in zip(paragraph._p.xpath("./w:hyperlink"), paragraph._p.xpath("./w:hyperlink/w:r")):
        hyperlink.addprevious(label_r)
        paragraph._p.remove(hyperlink)