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feature: Paragraph.text includes hyperlink text #85
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Hi Sebastian,
A .docx file is a ZIP archive, so it will make a lot more sense to you once you've unzipped it. If you're on a Mac or Linux this will be a good first step:
$ unzip -l some_document.docx
Contributors are always welcome, although I expect this project would be a steep learning curve for you. There is a related feature request here #74 you can take a look at and we can talk about what the API might look like for reading and writing hyperlinks if you're still interested.
Thanks for your reply, scanny. I've just read it right now and started researching. The command you posted revealed an inner file structure, so I used the Ubuntu's tool for compressed files and noticed that all of them are XML files. I'm used to XML through I made apps for Android and I used some SOAP webservices, so XML is not new for me. Althrough, you have a point about it may become a steep learning curve, through the XML structure seems to be quite complex.
Anyway, analyzing it I figured out some things in just less than half an hour: it seems that styles are defined in "styles.xml". There is also a file for the fonts, I just don't get why it seems there are 4 fonts in a test.docx file I created with LibreOffice in which I just used the default font and a hyperlink (which it seems it has it's own style defined), but I don't think extra fonts are relevant for now.
I've taken a look to the "document.xml" file and noticed a difference between a normal paragraph and a hyperlink paragraph: this would be a normal paragraph structure:
<w:p> <w:pPr> <w:pStyle w:val="style0"/> </w:pPr> <w:r> <w:rPr/> <w:t>Prueba jajajjajajajaja</w:t> </w:r> </w:p>On the other hand, this would be a hyperlink paragraph:
<w:p> <w:pPr> <w:pStyle w:val="style0"/> </w:pPr> <w:hyperlink r:id="rId2"> <w:r> <w:rPr> <w:rStyle w:val="style15"/> </w:rPr> <w:t>http://www.google.com/</w:t> </w:r> </w:hyperlink> </w:p>In other words, it seems that the tags <w:hyperlink></w:hyperlink> contain the whole rich text structure that is supposed to be the hyperlink, with an id which would point to the actual URL stored somewhere in the XML file system, I guess. It seems quite interesting, unfortunately, I don't have much spare time lately, because I'm very busy with web developing.
Anyways, if I ever have some spare time, I'd like to research how your python API reads the paragraphs, and make their .text() method able to recognize the <w:hyperlink> tag as text container. I'll keep you informed if I make any relevant progress.
Reacted by Georgy Chernousov and Fan, Qingxiang(Shirley)I think here's the problem: check the class CT_P at master/docx/oxml/text.py . If you take a look at the initial variables (lines 36 and 37) it seems this class (which I suppose it handles <w:p> objects) doesn't handle <w:hyperlink> objects at all. I think that's why text inside hyperlinks are not returned in the text property. I don't know much about the structure of the whole project -not yet-, but I think this is the way to go to resolve the problem.
This problem is very crucial for me, I hope the problem could be solved asap in the coming version.
Hi,
As indicated by SebasSBM above, the difference between a hyperlink and a paragraph is the [w:hyperlink r:id="XXX"] and [/w:hyperlink] tags. A workaround consists in removing those tags from the document's xml code, so that only the code of a standard paragraph remains (easy to do using regular expressions and the re module)
Here is how I did this:
I editedC:\Python27\Lib\site-packages\docx\oxml\__init__.pyas follows:1/ I created a new function that remove hyperlinks by nothing in a xml text:
def remove_hyperlink_tags(xml): import re xml = xml.replace("</w:hyperlink>","") xml = re.sub('<w:hyperlink[^>]*>',"",xml) return xml2/ I updated the standard parse_xml function as follows:
def parse_xml(xml): """ """ root_element = etree.fromstring(remove_hyperlink_tags(xml), oxml_parser) return root_elementIt worked well for me but I didn't test it much so use at your own risk...
Reacted by Georgy ChernousovHi,
python-docx is a great tool that i just discovered. However, i also faced the missing hyperlink text problem right away.
Please solve it.remove_hyperlink_tags don't work because i get:
File "C:\Python34\lib\site-packages\docx\oxml__init__.py", line 23, in remove_hyperlink_tags
xml = xml.replace("/w:hyperlink","")
TypeError: expected bytes, bytearray or buffer compatible objectOk, i think i faced the Python 2/3 issue. Here's something which works on Python 3.4:
def remove_hyperlink_tags(xml): import re text = xml.decode('utf-8') text = text.replace("</w:hyperlink>","") text = re.sub('<w:hyperlink[^>]*>', "", text) return text.encode('utf-8')
Reacted by Georgy Chernousov and thrinadhn- changed the title
[-]Getting hyperlinks from paragraphs[/-][+]feature: Paragraph.text includes hyperlink text[/+]on Feb 13, 2015 any schedule for that one?
- added a commit that references this issue
on Jun 13, 2016 I used the code at https://gist.github.com/etienned/7539105#file-extractdocx-py and supplied the path and get hyperlink texts as I need
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- added a commit that references this issue
on Sep 8, 2017 Until this functionality is implemented in python-docx, this is the workaround I used to redefine the
'text'property of thedocx.text.paragraph.Paragraphclass such that it includes hyperlinks.Necessary imports:
from docx.text.paragraph import Paragraph import reFirst I redefine the
textproperty with:Paragraph.text = property(lambda self: GetParagraphText(self))Every time
paragraph.textis called, the functionGetParagraphTextwill be called instead with the instanceparagraphof typedocx.text.paragraph.Paragraphas parameter.The function
GetParagraphTextis implemented as:def GetParagraphText(paragraph) def GetTag(element): return "%s:%s" % (element.prefix, re.match("{.*}(.*)", element.tag).group(1)) text = '' runCount = 0 for child in paragraph._p: tag = GetTag(child) if tag == "w:r": text += paragraph.runs[runCount].text runCount += 1 if tag == "w:hyperlink": for subChild in child: if GetTag(subChild) == "w:r": text += subChild.text return textThe above implementationis the least intrusive I could think of. It requires no modification to the rest of the code, does not need to create new objects, and re-uses when possible the logic already available in python-docx (e.g., paragraph.runs). Hopefully this will help!
Reacted by Mumpitz, Rachel, Massimo Gennaro, zouxunlong, Vadim Evard, Oleynikov Ivan, Mohamad Nour Almujarkesh, Thackery Archuletta, Matthias Simon, Billy Cao and 6 moreNice job @roydesbois :)
Following on from @roydesbois (thanks for the inspiration, a great solution), I needed the requirement to parse down to each individual run so styling and fonts an also be correctly interpreted.
This also uses the built in
qnto get the full qualified name of the tag, thus not requiring custom regex parsing. What's great about this is that calling.texton the paragraph still works as you'd expect, while allowing you to iterate through each run and grab its respective text/styles.from docx.oxml.shared import qn def GetParagraphRuns(paragraph): def _get(node, parent): for child in node: if child.tag == qn('w:r'): yield Run(child, parent) if child.tag == qn('w:hyperlink'): yield from _get(child, parent) return list(_get(paragraph._element, paragraph)) Paragraph.runs = property(lambda self: GetParagraphRuns(self))
Reacted by Olivier Roy, Massimo Gennaro, Kevin Pan, Paulo Ferreira, Margarita and uusamaReacted by APlusPeace, jWyb, Kevin Pan and Paulo Ferreira@tomking2's solution was great but in my case a lot of the links were embedded. "Click Here" was the text returned rather than the link "Click Here" related to. So I tweaked the approach to pull out the link from document.part.rels and added it to the child.text if it differed from the text (i.e. embedded) so "Click Here" becomes "Click Here"[https://www.google.com]
from docx.text.paragraph import Paragraph from docx.text.run import Run from docx.oxml.shared import qn def GetParagraphRuns(paragraph): def _get(node, parent, hyperlinkId=None): for child in node: if child.tag == qn('w:r'): if hyperlinkId: linkToAdd = document.part.rels[hyperlinkId]._target if child.text != linkToAdd: child.text = child.text + f'[{linkToAdd}]' yield Run(child, parent) if child.tag == qn('w:hyperlink'): hlid = child.attrib.get(qn('r:id')) yield from _get(child, parent, hlid) return list(_get(paragraph._element, paragraph)) Paragraph.runs = property(lambda self: GetParagraphRuns(self))
I've no doubt there is a better way to do this but it's worked a treat for me :)
Absolutely love JStooke's solution, but noticed a bug (with a super quick fix).
if child.text != linkToAdd:should beif linkToAdd not in child.text, as the hyperlink will never equal the text, and every time the method is called, it adds the hyperlink an additional time. That makes the full solution:from docx.text.paragraph import Paragraph from docx.text.run import Run from docx.oxml.shared import qn def GetParagraphRuns(paragraph): def _get(node, parent, hyperlinkId=None): for child in node: if child.tag == qn('w:r'): if hyperlinkId: linkToAdd = document.part.rels[hyperlinkId]._target if linkToAdd not in child.text: child.text = child.text + f'[{linkToAdd}]' yield Run(child, parent) if child.tag == qn('w:hyperlink'): hlid = child.attrib.get(qn('r:id')) yield from _get(child, parent, hlid) return list(_get(paragraph._element, paragraph)) Paragraph.runs = property(lambda self: GetParagraphRuns(self))
Also, the document variable is undefined within the method. I solved this by declaring it globally, e.g.
global document document = Document(file)
Reacted by JStooke, Isaac Kogan and Pavel ZubkoAwesome cheers @oliveslongjohns. I'd forgotten to mention i'd already defined the global variable earlier in my use case. Glad its working for you :D
@oliveslongjohns @JStooke thanks for absolutely great answer, it works very well. However, I'd like to ask if there is a way to adapt code to include in the paragraph.text return the hyperlink working, instead of text_link [ link ].
I found a function to create a hyperlink, but Idon't really know how to embed:`def add_hyperlink(paragraph, url, text, color, underline):
# This gets access to the document.xml.rels file and gets a new relation id value part = paragraph.part r_id = part.relate_to(url, docx.opc.constants.RELATIONSHIP_TYPE.HYPERLINK, is_external=True) # Create the w:hyperlink tag and add needed values hyperlink = docx.oxml.shared.OxmlElement('w:hyperlink') hyperlink.set(docx.oxml.shared.qn('r:id'), r_id, ) # Create a w:r element new_run = docx.oxml.shared.OxmlElement('w:r') # Create a new w:rPr element rPr = docx.oxml.shared.OxmlElement('w:rPr') # Add color if it is given if not color is None: c = docx.oxml.shared.OxmlElement('w:color') c.set(docx.oxml.shared.qn('w:val'), color) rPr.append(c) # Remove underlining if it is requested if not underline: u = docx.oxml.shared.OxmlElement('w:u') u.set(docx.oxml.shared.qn('w:val'), 'none') rPr.append(u) # Join all the xml elements together add add the required text to the w:r element new_run.append(rPr) new_run.text = text hyperlink.append(new_run) paragraph._p.append(hyperlink) return hyperlink`Added in v1.0.0 circa Oct 5, 2023.
When getting Document.paragraphs objects, their text method doesn't include hyperlinks in the output. The one with this problem posted his question here:
http://stackoverflow.com/questions/25228106/how-to-extract-text-from-an-existing-docx-file-using-python-docx/25228787#25228787
I've been reading the documentation of python-docx for several hours and didn't find any property or method useful to resolve this. Maybe some class should be created, or some methods should be appended to an existing class to achieve this.
I barely know something about python-docx API. I knew of it's existence trying to help some people in stackoverflow.com with their problems. I don't even know how Windows' DOCX format works (I tried to open it with an hexadecimal editor to try to figure it out, and I just don't get it :P). But I'm skilled with logic problems and I've got good skills with Python scripting. I'd like to help if there's something I can do.