Converting Doc file to HTML using python

Question

Below is my function which works fine when executed separately

import win32com.client
def document_tohtml(inputfile):
    print("Input File",inputfile)
    doc = win32com.client.GetObject(inputfile)
    saved_file = doc.SaveAs (FileName="C:\Extraction\TMS TO 692M15-22-F-00073 Monthly Status _ Financial Report July 2022.html", FileFormat=8)
    doc.Close ()
    return saved_file

if __name__ == '__main__':
    document_tohtml("C:\Extraction\TMS TO 692M15-22-F-00073 Monthly Status _ Financial Report July 2022.docx")

Issue : I am trying to convert the above function into Flask API which is below

@app.route('/doctohtml', methods=['GET','POST'])
def document_tohtml():
    if request.method == 'POST':
        files = request.files['file']
        with open(files.filename, "rb") as docx_file:
            print("Files",files)
            print("File name",files.filename)
        # print("Type",type(new_file))
            doc = win32com.client.GetObject(str(files))
        
        with open(files.filename, "w") as html_file:
            saved_file=html_file.write(doc.value)
        # saved_file = doc.SaveAs (FileName="sample.html", FileFormat=8)
        # doc.Close ()
    return saved_file

But when I am hitting the api via postman I am getting below error

Files 
File name TMS TO 692M15-22-F-00073 Monthly Status _ Financial Report July 2022.docx
127.0.0.1 - - [20/Sep/2022 16:11:06] "POST /doctohtml HTTP/1.1" 500 -
Traceback (most recent call last):
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 2548, in __call__
    return self.wsgi_app(environ, start_response)
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 2528, in wsgi_app
    response = self.handle_exception(e)
  File "C:\Extraction\Exenv\lib\site-packages\flask_cors\extension.py", line 165, in wrapped_function
    return cors_after_request(app.make_response(f(*args, **kwargs)))
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 2525, in wsgi_app
    response = self.full_dispatch_request()
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 1822, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "C:\Extraction\Exenv\lib\site-packages\flask_cors\extension.py", line 165, in wrapped_function
    return cors_after_request(app.make_response(f(*args, **kwargs)))
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 1820, in full_dispatch_request
    rv = self.dispatch_request()
  File "C:\Extraction\Exenv\lib\site-packages\flask\app.py", line 1796, in dispatch_request
    return self.ensure_sync(self.view_functions[rule.endpoint])(**view_args)
  **[![File "C:\Extraction\Doc_Extract
ew_approach.py", line 30, in document_tohtml
    doc = win32com.client.GetObject(str(files))
  File "C:\Extraction\Exenv\lib\site-packages\win32com\client\__init__.py", line 85, in GetObject
    return Moniker(Pathname, clsctx)
  File "C:\Extraction\Exenv\lib\site-packages\win32com\client\__init__.py", line 102, in Moniker
    moniker, i, bindCtx = pythoncom.MkParseDisplayName(Pathname)
pywintypes.com_error: (-2147221020, 'Invalid syntax', None, None)][1]][1]**

P.S I have also tried other libraries like mammoth,aspose but they are not providing desired results and then tried above pywin32

furas · Answer

I don't use Windows to test it but ...

... working code use filename in GetObject() but in flask you use str(file.files) which should give string like

and this is NOT filename

You may have to write data in file on disk (ie. using files.filename) and later use GetObject(files.filename)

Frankly, you should get error even for with open(files.filename, "rb") beause this file still exists only in memory (in Python's variables) but not as file on disk - so you can't open it.

Converting Doc file to HTML using python

1 Answers