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How do you correctly parse web links to avoid a 403 error when using Wget?

I just started learning python yesterday and have VERY minimal coding skill. I am trying to write a python script that will process a folder of PDFs. Each PDF contains at least 1, and maybe as many as 15 or more, web links to supplemental documents. I think I'm off to a good start, but I'm having consistent "HTTP Error 403: Forbidden" errors when trying to use the wget function. I believe I'm just not parsing the web links correctly. I think the main issue is coming in because the web links are mostly "s3.amazonaws.com" links that are SUPER long.
For reference:
Link copied directly from PDF (works to download): https://s3.amazonaws.com/os_uploads/2169504_DFA%20train%20pass.PNG?AWSAccessKeyId=AKIAIPCTK7BDMEW7SP4Q&Expires=1909634500&Signature=aQlQXVR8UuYLtkzjvcKJ5tiVrZQ=&response-content-disposition=attachment;%20filename*=utf-8''DFA%2520train%2520pass.PNG
Link as it appears after trying to parse it in my code (doesn't work, gives "unknown url type" when trying to download): https%3A//s3.amazonaws.com/os_uploads/2169504_DFA%2520train%2520pass.PNG%3FAWSAccessKeyId%3DAKIAIPCTK7BDMEW7SP4Q%26Expires%3D1909634500%26Signature%3DaQlQXVR8UuYLtkzjvcKJ5tiVrZQ%253D%26response-content-disposition%3Dattachment%253B%2520filename%252A%253Dutf-8%2527%2527DFA%252520train%252520pass.PNG
Additionally if people want to weigh in on how I'm doing this in a stupid way. Each PDF starts with a string of 6 digits, and once I download supplemental documents I want to auto save and name them as XXXXXX_attachY.* Where X is the identifying string of digits and Y just increases for each attachment. I haven't gotten my code to work enough to test that, but I'm fairly certain I don't have it correct either.
Help!
#!/usr/bin/env python3
import os
import glob
import pdfx
import wget
import urllib.parse
## Accessing and Creating Six Digit File Code
pdf_dir = "/users/USERNAME/desktop/worky"
pdf_files = glob.glob("%s/*.pdf" % pdf_dir)
for file in pdf_files:
## Identify File Name and Limit to Digits
filename = os.path.basename(file)
newname = filename[0:6]
## Run PDFX to identify and download links
pdf = pdfx.PDFx(filename)
url_list = pdf.get_references_as_dict()
attachment_counter = (1)
for x in url_list["url"]:
if x[0:4] == "http":
parsed_url = urllib.parse.quote(x, safe='://')
print (parsed_url)
wget.download(parsed_url, '/users/USERNAME/desktop/worky/(newname)_attach(attachment_counter).*')
##os.rename(r'/users/USERNAME/desktop/worky/(filename).*',r'/users/USERNAME/desktop/worky/(newname)_attach(attachment_counter).*')
attachment_counter += 1
for x in url_list["pdf"]:
print (parsed_url + "\n")```

I prefer to use requests (https://requests.readthedocs.io/en/master/) when trying to grab text or files online. I tried it quickly with wget and I got the same error (might be linked to user-agent HTTP headers used by wget).
wget and HTTP headers issues : download image from url using python urllib but receiving HTTP Error 403: Forbidden
HTTP headers : https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/User-Agent
The good thing with requests is that it lets you modify HTTP headers the way you want (https://requests.readthedocs.io/en/master/user/quickstart/#custom-headers).
import requests
r = requests.get("https://s3.amazonaws.com/os_uploads/2169504_DFA%20train%20pass.PNG?AWSAccessKeyId=AKIAIPCTK7BDMEW7SP4Q&Expires=1909634500&Signature=aQlQXVR8UuYLtkzjvcKJ5tiVrZQ=&response-content-disposition=attachment;%20filename*=utf-8''DFA%2520train%2520pass.PNG")
with open("myfile.png", "wb") as file:
file.write(r.content)
I'm not sure I understand what you're trying to do, but maybe you want to use formatted strings to build your URLs (https://docs.python.org/3/library/stdtypes.html?highlight=format#str.format) ?
Maybe checking string indexes is fine in your case (if x[0:4] == "http":), but I think you should check python re package to use regular expressions to catch the elements you want in a document (https://docs.python.org/3/library/re.html).
import re
regex = re.compile(r"^http://")
if re.match(regex, mydocument):
<do something>

The reason for this behavior is inside wget library. Inside it encodes the URL with urllib.parse.quote() (https://docs.python.org/3/library/urllib.parse.html#urllib.parse.quote).
Basically it replaces characters with their appropriate %xx escape character. Your URL is already escaped but the library does not know that. When it parses the %20 it sees % as a character that needs to be replaced so the result is %2520 and different URL - therefore 403 error.
You could decode that URL first and then pass it, but then you would have another problem with this library because your URL has parameter filename*= but the library expects filename=.
I would recommend doing something like this:
# get the file
req = requests.get(parsed_url)
# parse your URL to get GET parameters
get_parameters = [x for x in parsed_url.split('?')[1].split('&')]
filename = ''
# find the get parameter with the name
for get_parameter in get_parameters:
if "filename*=" in get_parameter:
# split it to get the name
filename = get_parameter.split('filename*=')[1]
# save the file
with open(<path> + filename, 'wb') as file:
file.write(req.content)
I would also recommend removing the utf-8'' in that filename because I don't think it is actually part of the filename. You could also use regular expressions for getting the filename, but this was easier for me.

urllib2 download HTML file

Using urllib2 in Python 2.7.4, I can readily download an Excel file:
output_file = 'excel.xls'
url = 'http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
This results in the expected file that I can use as I wish.
However, trying to download just an HTML file gives me an empty file:
output_file = 'webpage.html'
url = 'http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
I had the same results using urllib. There must be something simple I'm missing or don't understand. How do I download an HTML file from a URL? Why doesn't my code work?

If you want to download files or simply save a webpage you can use urlretrieve(from urllib library)instead of use read and write.
import urllib
urllib.urlretrieve("http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html","doc.html")
#urllib.urlretrieve("url","save as..")
If you need to set a timeout you have to put it at the start of your file:
import socket
socket.setdefaulttimeout(25)
#seconds

It also Python 2.7.4 in my OS X 10.9, and the codes work well on it.
So I think there maybe other problems prevent its working. Can you open "http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls" in your browser?

This may not directly answer the question, but if you're working with HTTP and have sufficient privileges to install python packages, I'd really recommend doing this with 'requests'. There's a related answered here - https://stackoverflow.com/a/13137873/45698

Pyquery： I'm using pyquery ,the HTTP_REFFER is required in this page,How can i do with it?

the website needs HTTP_REFFER when i send request..
the common way to open pages in PyQuery is `
> doc=pyQuery(url=r'http://www.....')
how can i add HTTP_REFFER ?
`

pyQuery uses urlopen from urllib.request if you're on py3 or urllib2 if you're on py2. When you feed it with the url parameter it should either be a string or a Request object.
In the python2 case let's see how it would look like if you want to add an http_header to your request:
import urllib2
url = urllib2.Request("http://...", headers={'HTTP_REFERER': "http://..."})
doc = pyQuery(url=url)
It would be similar in the python3 case. It's always good to read through the code of the libs your're working with, you can find the pyQuery code here.

Any way of getting PyDoc into Jira Confluence

I'm using PyDoc to generate documentation from my Python code and I'm using Jira's Confluence plugin to manage documentation. Is there any way to generating PyDoc documentation and putting it into Confluence?
Googling didn't yield too many results.
Thanks everyone

You can try something like this:
from pydoc import *
import io
d = HTMLDoc()
content = d.docmodule(sys.modules["mymodule"])
f = io.open('./out.html', 'w')
f.write(unicode(content))
f.close()
You now have an HTML file containing the pydoc info. The next trick is to get it into Confluence so that it looks nice. I have so far tried importing it into Microsoft Word as an .rtf, then cut-and-paste into Confluence.

pydoc generates one file format, html. So your challenge is getting an self contained HTML4 page into Confluence.
The Confluence wiki says you can use a HTML macro.
I imagine ideally you'd want to get pydoc to generate Confluence Markup, but you'd need a different tool for that.

Python urllib2 file upload problems

I'm currently trying to initiate a file upload with urllib2 and the urllib2_file library. Here's my code:
import sys
import urllib2_file
import urllib2
URL='http://aquate.us/upload.php'
d = [('uploaded', open(sys.argv[1:]))]
req = urllib2.Request(URL, d)
u = urllib2.urlopen(req)
print u.read()
I've placed this .py file in my My Documents directory and placed a shortcut to it in my Send To folder (the shortcut URL is ).
When I right click a file, choose Send To, and select Aquate (my python), it opens a command prompt for a split second and then closes it. Nothing gets uploaded.
I knew there was probably an error going on so I typed the code into CL python, line by line.
When I ran the u=urllib2.urlopen(req) line, I didn't get an error;
alt text http://www.aquate.us/u/55245858877937182052.jpg
instead, the cursor simply started blinking on a new line beneath that line. I waited a couple of minutes to see if something would happen but it just stayed like that. To get it to stop, I had to press ctrl+break.
What's up with this script?
Thanks in advance!
[Edit]
Forgot to mention -- when I ran the script without the request data (the file) it ran like a charm. Is it a problem with urllib2_file?
[edit 2]:
import MultipartPostHandler, urllib2, cookielib,sys
import win32clipboard as w
cookies = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookies),MultipartPostHandler.MultipartPostHandler)
params = {"uploaded" : open("c:/cfoot.js") }
a=opener.open("http://www.aquate.us/upload.php", params)
text = a.read()
w.OpenClipboard()
w.EmptyClipboard()
w.SetClipboardText(text)
w.CloseClipboard()
That code works like a charm if you run it through the command line.

If you're using Python 2.5 or newer, urllib2_file is both unnecessary and unsupported, so check which version you're using (and perhaps upgrade).
If you're using Python 2.3 or 2.4 (the only versions supported by urllib2_file), try running the sample code and see if you have the same problem. If so, there is likely something wrong either with your Python or urllib2_file installation.
EDIT:
Also, you don't seem to be using either of urllib2_file's two supported formats for POST data. Try using one of the following two lines instead:
d = ['uploaded', open(sys.argv[1:])]
## --OR-- ##
d = {'uploaded': open(sys.argv[1:])}

First, there's a third way to run Python programs.
From cmd.exe, type python myprogram.py. You get a nice log. You don't have to type stuff one line at a time.
Second, check the urrlib2 documentation. You'll need to look at urllib, also.
A Request requires a URL and a urlencoded encoded buffer of data.
data should be a buffer in the
standard
application/x-www-form-urlencoded
format. The urllib.urlencode()
function takes a mapping or sequence
of 2-tuples and returns a string in
this format.
You need to encode your data.

If you're still on Python2.5, what worked for me was to download the code here:
http://peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html
and save it as MultipartPostHandler.py
then use:
import urllib2, MultipartPostHandler
opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})
or if you need cookies:
import urllib2, MultipartPostHandler, cookielib
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj), MultipartPostHandler.MultipartPostHandler())
opener.open(url, {"file":open(...)})