I'am tasked with converting tons of .doc files to .pdf. And the only way my supervisor wants me to do this is through MSWord 2010. I know I should be able to automate this with python COM automation. Only problem is I dont know how and where to start. I tried searching for some tutorials but was not able to find any (May be I might have, but I don't know what I'm looking for).
Right now I'm reading through this. Dont know how useful this is going to be.
A simple example using comtypes, converting a single file, input and output filenames given as commandline arguments:
import sys
import os
import comtypes.client
wdFormatPDF = 17
in_file = os.path.abspath(sys.argv[1])
out_file = os.path.abspath(sys.argv[2])
word = comtypes.client.CreateObject('Word.Application')
doc = word.Documents.Open(in_file)
doc.SaveAs(out_file, FileFormat=wdFormatPDF)
doc.Close()
word.Quit()
You could also use pywin32, which would be the same except for:
import win32com.client
and then:
word = win32com.client.Dispatch('Word.Application')
You can use the docx2pdf python package to bulk convert docx to pdf. It can be used as both a CLI and a python library. It requires Microsoft Office to be installed and uses COM on Windows and AppleScript (JXA) on macOS.
from docx2pdf import convert
convert("input.docx")
convert("input.docx", "output.pdf")
convert("my_docx_folder/")
pip install docx2pdf
docx2pdf input.docx output.pdf
Disclaimer: I wrote the docx2pdf package. https://github.com/AlJohri/docx2pdf
I have tested many solutions but no one of them works efficiently on Linux distribution.
I recommend this solution :
import sys
import subprocess
import re
def convert_to(folder, source, timeout=None):
args = [libreoffice_exec(), '--headless', '--convert-to', 'pdf', '--outdir', folder, source]
process = subprocess.run(args, stdout=subprocess.PIPE, stderr=subprocess.PIPE, timeout=timeout)
filename = re.search('-> (.*?) using filter', process.stdout.decode())
return filename.group(1)
def libreoffice_exec():
# TODO: Provide support for more platforms
if sys.platform == 'darwin':
return '/Applications/LibreOffice.app/Contents/MacOS/soffice'
return 'libreoffice'
and you call your function:
result = convert_to('TEMP Directory', 'Your File', timeout=15)
All resources:
https://michalzalecki.com/converting-docx-to-pdf-using-python/
I have worked on this problem for half a day, so I think I should share some of my experience on this matter. Steven's answer is right, but it will fail on my computer. There are two key points to fix it here:
(1). The first time when I created the 'Word.Application' object, I should make it (the word app) visible before open any documents. (Actually, even I myself cannot explain why this works. If I do not do this on my computer, the program will crash when I try to open a document in the invisible model, then the 'Word.Application' object will be deleted by OS. )
(2). After doing (1), the program will work well sometimes but may fail often. The crash error "COMError: (-2147418111, 'Call was rejected by callee.', (None, None, None, 0, None))" means that the COM Server may not be able to response so quickly. So I add a delay before I tried to open a document.
After doing these two steps, the program will work perfectly with no failure anymore. The demo code is as below. If you have encountered the same problems, try to follow these two steps. Hope it helps.
import os
import comtypes.client
import time
wdFormatPDF = 17
# absolute path is needed
# be careful about the slash '\', use '\\' or '/' or raw string r"..."
in_file=r'absolute path of input docx file 1'
out_file=r'absolute path of output pdf file 1'
in_file2=r'absolute path of input docx file 2'
out_file2=r'absolute path of outputpdf file 2'
# print out filenames
print in_file
print out_file
print in_file2
print out_file2
# create COM object
word = comtypes.client.CreateObject('Word.Application')
# key point 1: make word visible before open a new document
word.Visible = True
# key point 2: wait for the COM Server to prepare well.
time.sleep(3)
# convert docx file 1 to pdf file 1
doc=word.Documents.Open(in_file) # open docx file 1
doc.SaveAs(out_file, FileFormat=wdFormatPDF) # conversion
doc.Close() # close docx file 1
word.Visible = False
# convert docx file 2 to pdf file 2
doc = word.Documents.Open(in_file2) # open docx file 2
doc.SaveAs(out_file2, FileFormat=wdFormatPDF) # conversion
doc.Close() # close docx file 2
word.Quit() # close Word Application
unoconv (writen in Python) and OpenOffice running as a headless daemon.
https://github.com/unoconv/unoconv
http://dag.wiee.rs/home-made/unoconv/
Works very nicely for doc, docx, ppt, pptx, xls, xlsx.
Very useful if you need to convert docs or save/convert to certain formats on a server.
As an alternative to the SaveAs function, you could also use ExportAsFixedFormat which gives you access to the PDF options dialog you would normally see in Word. With this you can specify bookmarks and other document properties.
doc.ExportAsFixedFormat(OutputFileName=pdf_file,
ExportFormat=17, #17 = PDF output, 18=XPS output
OpenAfterExport=False,
OptimizeFor=0, #0=Print (higher res), 1=Screen (lower res)
CreateBookmarks=1, #0=No bookmarks, 1=Heading bookmarks only, 2=bookmarks match word bookmarks
DocStructureTags=True
);
The full list of function arguments is: 'OutputFileName', 'ExportFormat', 'OpenAfterExport', 'OptimizeFor', 'Range', 'From', 'To', 'Item', 'IncludeDocProps', 'KeepIRM', 'CreateBookmarks', 'DocStructureTags', 'BitmapMissingFonts', 'UseISO19005_1', 'FixedFormatExtClassPtr'
It's worth noting that Stevens answer works, but make sure if using a for loop to export multiple files to place the ClientObject or Dispatch statements before the loop - it only needs to be created once - see my problem: Python win32com.client.Dispatch looping through Word documents and export to PDF; fails when next loop occurs
If you don't mind using PowerShell have a look at this Hey, Scripting Guy! article. The code presented could be adopted to use the wdFormatPDF enumeration value of WdSaveFormat (see here).
This blog article presents a different implementation of the same idea.
I have modified it for ppt support as well. My solution support all the below-specified extensions.
word_extensions = [".doc", ".odt", ".rtf", ".docx", ".dotm", ".docm"]
ppt_extensions = [".ppt", ".pptx"]
My Solution: Github Link
I have modified code from Docx2PDF
I tried the accepted answer but wasn't particularly keen on the bloated PDFs Word was producing which was usually an order of magnitude bigger than expected. After looking how to disable the dialogs when using a virtual PDF printer I came across Bullzip PDF Printer and I've been rather impressed with its features. It's now replaced the other virtual printers I used previously. You'll find a "free community edition" on their download page.
The COM API can be found here and a list of the usable settings can be found here. The settings are written to a "runonce" file which is used for one print job only and then removed automatically. When printing multiple PDFs we need to make sure one print job completes before starting another to ensure the settings are used correctly for each file.
import os, re, time, datetime, win32com.client
def print_to_Bullzip(file):
util = win32com.client.Dispatch("Bullzip.PDFUtil")
settings = win32com.client.Dispatch("Bullzip.PDFSettings")
settings.PrinterName = util.DefaultPrinterName # make sure we're controlling the right PDF printer
outputFile = re.sub("\.[^.]+$", ".pdf", file)
statusFile = re.sub("\.[^.]+$", ".status", file)
settings.SetValue("Output", outputFile)
settings.SetValue("ConfirmOverwrite", "no")
settings.SetValue("ShowSaveAS", "never")
settings.SetValue("ShowSettings", "never")
settings.SetValue("ShowPDF", "no")
settings.SetValue("ShowProgress", "no")
settings.SetValue("ShowProgressFinished", "no") # disable balloon tip
settings.SetValue("StatusFile", statusFile) # created after print job
settings.WriteSettings(True) # write settings to the runonce.ini
util.PrintFile(file, util.DefaultPrinterName) # send to Bullzip virtual printer
# wait until print job completes before continuing
# otherwise settings for the next job may not be used
timestamp = datetime.datetime.now()
while( (datetime.datetime.now() - timestamp).seconds < 10):
if os.path.exists(statusFile) and os.path.isfile(statusFile):
error = util.ReadIniString(statusFile, "Status", "Errors", '')
if error != "0":
raise IOError("PDF was created with errors")
os.remove(statusFile)
return
time.sleep(0.1)
raise IOError("PDF creation timed out")
I was working with this solution but I needed to search all .docx, .dotm, .docm, .odt, .doc or .rtf and then turn them all to .pdf (python 3.7.5). Hope it works...
import os
import win32com.client
wdFormatPDF = 17
for root, dirs, files in os.walk(r'your directory here'):
for f in files:
if f.endswith(".doc") or f.endswith(".odt") or f.endswith(".rtf"):
try:
print(f)
in_file=os.path.join(root,f)
word = win32com.client.Dispatch('Word.Application')
word.Visible = False
doc = word.Documents.Open(in_file)
doc.SaveAs(os.path.join(root,f[:-4]), FileFormat=wdFormatPDF)
doc.Close()
word.Quit()
word.Visible = True
print ('done')
os.remove(os.path.join(root,f))
pass
except:
print('could not open')
# os.remove(os.path.join(root,f))
elif f.endswith(".docx") or f.endswith(".dotm") or f.endswith(".docm"):
try:
print(f)
in_file=os.path.join(root,f)
word = win32com.client.Dispatch('Word.Application')
word.Visible = False
doc = word.Documents.Open(in_file)
doc.SaveAs(os.path.join(root,f[:-5]), FileFormat=wdFormatPDF)
doc.Close()
word.Quit()
word.Visible = True
print ('done')
os.remove(os.path.join(root,f))
pass
except:
print('could not open')
# os.remove(os.path.join(root,f))
else:
pass
The try and except was for those documents I couldn't read and won't exit the code until the last document.
You should start from investigating so called virtual PDF print drivers.
As soon as you will find one you should be able to write batch file that prints your DOC files into PDF files. You probably can do this in Python too (setup printer driver output and issue document/print command in MSWord, later can be done using command line AFAIR).
import docx2txt
from win32com import client
import os
files_from_folder = r"c:\\doc"
directory = os.fsencode(files_from_folder)
amount = 1
word = client.DispatchEx("Word.Application")
word.Visible = True
for file in os.listdir(directory):
filename = os.fsdecode(file)
print(filename)
if filename.endswith('docx'):
text = docx2txt.process(os.path.join(files_from_folder, filename))
print(f'{filename} transfered ({amount})')
amount += 1
new_filename = filename.split('.')[0] + '.txt'
try:
with open(os.path.join(files_from_folder + r'\txt_files', new_filename), 'w', encoding='utf-8') as t:
t.write(text)
except:
os.mkdir(files_from_folder + r'\txt_files')
with open(os.path.join(files_from_folder + r'\txt_files', new_filename), 'w', encoding='utf-8') as t:
t.write(text)
elif filename.endswith('doc'):
doc = word.Documents.Open(os.path.join(files_from_folder, filename))
text = doc.Range().Text
doc.Close()
print(f'{filename} transfered ({amount})')
amount += 1
new_filename = filename.split('.')[0] + '.txt'
try:
with open(os.path.join(files_from_folder + r'\txt_files', new_filename), 'w', encoding='utf-8') as t:
t.write(text)
except:
os.mkdir(files_from_folder + r'\txt_files')
with open(os.path.join(files_from_folder + r'\txt_files', new_filename), 'w', encoding='utf-8') as t:
t.write(text)
word.Quit()
The Source Code, see here:
https://neculaifantanaru.com/en/python-full-code-how-to-convert-doc-and-docx-files-to-pdf-from-the-folder.html
I would suggest ignoring your supervisor and use OpenOffice which has a Python api. OpenOffice has built in support for Python and someone created a library specific for this purpose (PyODConverter).
If he isn't happy with the output, tell him it could take you weeks to do it with word.
I have this script which download all images from a given web url address:
from selenium import webdriver
import urllib
class ChromefoxTest:
def __init__(self,url):
self.url=url
self.uri = []
def chromeTest(self):
# file_name = "C:\Users\Administrator\Downloads\images"
self.driver=webdriver.Chrome()
self.driver.get(self.url)
self.r=self.driver.find_elements_by_tag_name('img')
# output=open(file_name,'w')
for i, v in enumerate(self.r):
src = v.get_attribute("src")
self.uri.append(src)
pos = len(src) - src[::-1].index('/')
print src[pos:]
self.g=urllib.urlretrieve(src, src[pos:])
# output.write(src)
# output.close()
if __name__=='__main__':
FT=ChromefoxTest("http://imgur.com/")
FT.chromeTest()
my question is: how do i make this script to save all the pics to a specific folder location on my windows machine?
You need to specify the path where you want to save the file. This is explained in the documentation for urllib.urlretrieve:
The method is: urllib.urlretrieve(url[, filename[, reporthook[, data]]]).
And the documentation says:
The second argument, if present, specifies the file location to copy to (if absent, the location will be a tempfile with a generated name).
So...
urllib.urlretrieve(src, 'location/on/my/system/foo.png')
Will save the image to the specified folder.
Also, consider reviewing the documentation for os.path. Those functions will help you manipulate file names and paths.
If you use the requests library you can slurp up really big image files (or small ones) efficiently and arrange to store them in a place of your choice in an obvious way.
Use this code and you'll get a nice picture of a beagle dog!
image_url is the link to the remote image.
file_path is where you want to store the image locally. It can include just a file name or a full path, at your option.
chunk_size is the size of the piece of the file to be downloaded with each slurp from the remote site.
length is the actual size of the piece that is written locally. Since I did this interactively I put this in mainly so that I wouldn't have to look at a long vertical stream of 1024s on my screen.
..
>>> import requests
>>> image_url = 'http://maxpixel.freegreatpicture.com/static/photo/1x/Eyes-Dog-Portrait-Animal-Familiar-Domestic-Beagle-2507963.jpg'
>>> file_path = r'c:\scratch\beagle.jpg'
>>> r = requests.get(image_url, stream=True)
>>> with open(file_path, 'wb') as beagle:
... for chunk in r.iter_content(chunk_size=1024):
... length = beagle.write(chunk)
I'm trying to save a bunch of pages in a folder next to the py file that creates them. I'm on windows so when I try to make the trailing backslash for the file-path it makes a special character instead.
Here's what I'm talking about:
from bs4 import BeautifulSoup
import urllib2, urllib
import csv
import requests
from os.path import expanduser
print "yes"
with open('intjpages.csv', 'rb') as csvfile:
pagereader = csv.reader(open("intjpages.csv","rb"))
i=0
for row in pagereader:
print row
agentheader = {'User-Agent': 'Nerd'}
request = urllib2.Request(row[0],headers=agentheader)
url = urllib2.urlopen(request)
soup = BeautifulSoup(url)
for div in soup.findAll('div', {"class" : "side"}):
div.extract()
body = soup.find_all("div", { "class" : "md" })
name = "page" + str(i) + ".html"
path_to_file = "\cleanishdata\"
outfile = open(path_to_file + name, 'w')
#outfile = open(name,'w') #this works fine
body=str(body)
outfile.write(body)
outfile.close()
i+=1
I can save the files to the same folder that the .py file is in, but when I process the files using rapidminer it includes the program too. Also it would just be neater if I could save it in a directory.
I am surprised this hasn't already been answered on the entire internet.
EDIT: Thanks so much! I ended up using information from both of your answers. IDLE was making me use r'\string\' to concatenate the strings with the backslashes. I needed use the path_to_script technique of abamert to solve the problem of creating a new folder wherever the py file is. Thanks again! Here's the relevant coding changes:
name = "page" + str(i) + ".txt"
path_to_script_dir = os.path.dirname(os.path.abspath("links.py"))
newpath = path_to_script_dir + r'\\' + 'cleanishdata'
if not os.path.exists(newpath): os.makedirs(newpath)
outfile = open(path_to_script_dir + r'\\cleanishdata\\' + name, 'w')
body=str(body)
outfile.write(body)
outfile.close()
i+=1
Are you sure sure you're escaping your backslashes properly?
The \" in your string "\cleanishdata\" is actually an escaped quote character (").
You probably want
r"\cleanishdata\"
or
"\\cleanishdata\\"
You probably also want to check out the os.path library, particular os.path.join and os.path.dirname.
For example, if your file is in C:\Base\myfile.py and you want to save files to C:\Base\cleanishdata\output.txt, you'd want:
os.path.join(
os.path.dirname(os.path.abspath(sys.argv[0])), # C:\Base\
'cleanishdata',
'output.txt')
A better solution than hardcoding the path to the .py file is to just ask Python for it:
import sys
import os
path_to_script = sys.argv[0]
path_to_script_dir = os.path.dirname(os.path.abspath(path_to_script))
Also, it's generally better to use os.path methods instead of string manipulation:
outfile = open(os.path.join(path_to_script_dir, name), 'w')
Besides making your program continue to work as expected even if you move it to a different location or install it on another machine or give it to a friend, getting rid of the hardcoded paths and the string-based path concatenation means you don't have to worry about backslashes anywhere, and this problem never arises in the first place.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Serving dynamically generated ZIP archives in Django
(Feel free to point me to any potential duplicates if I have missed them)
I have looked at this snippet:
http://djangosnippets.org/snippets/365/
and this answer:
but I wonder how I can tweak them to suit my need: I want multiple files to be zipped and the archive available as a download via a link (or dynamically generated via a view). I am new to Python and Django so I don't know how to go about it.
Thank in advance!
I've posted this on the duplicate question which Willy linked to, but since questions with a bounty cannot be closed as a duplicate, might as well copy it here too:
import os
import zipfile
import StringIO
from django.http import HttpResponse
def getfiles(request):
# Files (local path) to put in the .zip
# FIXME: Change this (get paths from DB etc)
filenames = ["/tmp/file1.txt", "/tmp/file2.txt"]
# Folder name in ZIP archive which contains the above files
# E.g [thearchive.zip]/somefiles/file2.txt
# FIXME: Set this to something better
zip_subdir = "somefiles"
zip_filename = "%s.zip" % zip_subdir
# Open StringIO to grab in-memory ZIP contents
s = StringIO.StringIO()
# The zip compressor
zf = zipfile.ZipFile(s, "w")
for fpath in filenames:
# Calculate path for file in zip
fdir, fname = os.path.split(fpath)
zip_path = os.path.join(zip_subdir, fname)
# Add file, at correct path
zf.write(fpath, zip_path)
# Must close zip for all contents to be written
zf.close()
# Grab ZIP file from in-memory, make response with correct MIME-type
resp = HttpResponse(s.getvalue(), mimetype = "application/x-zip-compressed")
# ..and correct content-disposition
resp['Content-Disposition'] = 'attachment; filename=%s' % zip_filename
return resp
So as I understand your problem is not how to generate dynamically this file, but creating a link for people to download it...
What I suggest is the following:
0) Create a model for your file, if you want to generate it dynamically don't use the FileField, but just the info you need for generating this file:
class ZipStored(models.Model):
zip = FileField(upload_to="/choose/a/path/")
1) Create and store your Zip. This step is important, you create your zip in memory, and then cast it to assign it to the FileField:
function create_my_zip(request, [...]):
[...]
# This is a in-memory file
file_like = StringIO.StringIO()
# Create your zip, do all your stuff
zf = zipfile.ZipFile(file_like, mode='w')
[...]
# Your zip is saved in this "file"
zf.close()
file_like.seek(0)
# To store it we can use a InMemoryUploadedFile
inMemory = InMemoryUploadedFile(file_like, None, "my_zip_%s" % filename, 'application/zip', file_like.len, None)
zip = ZipStored(zip=inMemory)
# Your zip will be stored!
zip.save()
# Notify the user the zip was created or whatever
[...]
2) Create a url, for example get a number matching the id, you can also use a slugfield (this)
url(r'^get_my_zip/(\d+)$', "zippyApp.views.get_zip")
3) Now the view, this view will return the file matching the id passed in the url, you can also use a slug sending the text instead of the id, and make the get filtering by your slugfield.
function get_zip(request, id):
myzip = ZipStored.object.get(pk = id)
filename = myzip.zip.name.split('/')[-1]
# You got the zip! Now, return it!
response = HttpResponse(myzip.file, content_type='application/zip')
response['Content-Disposition'] = 'attachment; filename=%s' % filename
I am trying to download a zip file to a local drive and extract all files to a destination folder.
so i have come up with solution but it is only to "download" a file from a directory to another directory but it doesn't work for downloading files. for the extraction, I am able to get it to work in 2.6 but not for 2.5. so any suggestions for the work around or another approach I am definitely open to.
thanks in advance.
######################################
'''this part works but it is not good for URl links'''
import shutil
sourceFile = r"C:\Users\blueman\master\test2.5.zip"
destDir = r"C:\Users\blueman\user"
shutil.copy(sourceFile, destDir)
print "file copied"
######################################################
'''extract works but not good for version 2.5'''
import zipfile
GLBzipFilePath =r'C:\Users\blueman\user\test2.5.zip'
GLBextractDir =r'C:\Users\blueman\user'
def extract(zipFilePath, extractDir):
zip = zipfile(zipFilePath)
zip.extractall(path=extractDir)
print "it works"
extract(GLBzipFilePath,GLBextractDir)
######################################################
urllib.urlretrieve can get a file (zip or otherwise;-) from a URL to a given path.
extractall is indeed new in 2.6, but in 2.5 you can use an explicit loop (get all names, open each name, etc). Do you need example code?
So here's the general idea (needs more try/except if you want to give a nice error message in each and every case which could go wrong, of which, of course, there are a million variants -- I'm only using a couple of such cases as examples...):
import os
import urllib
import zipfile
def getunzipped(theurl, thedir):
name = os.path.join(thedir, 'temp.zip')
try:
name, hdrs = urllib.urlretrieve(theurl, name)
except IOError, e:
print "Can't retrieve %r to %r: %s" % (theurl, thedir, e)
return
try:
z = zipfile.ZipFile(name)
except zipfile.error, e:
print "Bad zipfile (from %r): %s" % (theurl, e)
return
for n in z.namelist():
dest = os.path.join(thedir, n)
destdir = os.path.dirname(dest)
if not os.path.isdir(destdir):
os.makedirs(destdir)
data = z.read(n)
f = open(dest, 'w')
f.write(data)
f.close()
z.close()
os.unlink(name)
For downloading, look at urllib:
import urllib
webFile = urllib.urlopen(url)
For unzipping, use zipfile. See also this example.
The shortest way i've found so far, is to use +alex answer, but with ZipFile.extractall() instead of the loop:
from zipfile import ZipFile
from urllib import urlretrieve
from tempfile import mktemp
filename = mktemp('.zip')
destDir = mktemp()
theurl = 'http://www.example.com/file.zip'
name, hdrs = urlretrieve(theurl, filename)
thefile=ZipFile(filename)
thefile.extractall(destDir)
thefile.close()