AttributeError: 'module' object has no attribute 'urlopen'

I'm trying to use Python to download the HTML source code of a website but I'm receiving this error.

Traceback (most recent call last): File "C:\Users\Sergio.Tapia\Documents\NetBeansProjects\DICParser\src\WebDownload.py", line 3, in <module> file = urllib.urlopen("") AttributeError: 'module' object has no attribute 'urlopen' 

I'm following the guide here:

import urllib file = urllib.urlopen("") s = file.read() f.close() #I'm guessing this would output the html source code? print(s) 

I'm using Python 3.

13 Answers

This works in Python 2.x.

For Python 3 look in the docs:

import urllib.request with urllib.request.urlopen("") as url: s = url.read() # I'm guessing this would output the html source code ? print(s) 

A Python 2+3 compatible solution is:

import sys if sys.version_info[0] == 3: from urllib.request import urlopen else: # Not Python 3 - today, it is most likely to be Python 2 # But note that this might need an update when Python 4 # might be around one day from urllib import urlopen # Your code where you can use urlopen with urlopen("") as url: s = url.read() print(s) 

import urllib.request as ur s = ur.urlopen("") sl = s.read() print(sl) 

In Python v3 the "urllib.request" is a module by itself, therefore "urllib" cannot be used here.

To get 'dataX = urllib.urlopen(url).read()' working in python3 (this would have been correct for python2) you must just change 2 little things.

1: The urllib statement itself (add the .request in the middle):

dataX = urllib.request.urlopen(url).read() 

2: The import statement preceding it (change from 'import urlib' to:

import urllib.request 

And it should work in python3 :)

Change TWO lines:

import urllib.request #line1 #Replace urllib.urlopen("") #To urllib.request.urlopen("") #line2 

If You got ERROR 403: Forbidden Error exception try this:

siteurl = "" req = urllib.request.Request(siteurl, headers={'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/80.0.3987.100 Safari/537.36'}) pageHTML = urllib.request.urlopen(req).read() 

I hope your problem resolved.

import urllib.request as ur filehandler = ur.urlopen (') for line in filehandler: print(line.strip()) 

For python 3, try something like this:

import urllib.request urllib.request.urlretrieve(' "video_name.avi") 

It will download the video to the current working directory

I got help from HERE

Solution for python3:

from urllib.request import urlopen url = ' file = urlopen(url) html = file.read() print(html) 

import urllib import urllib.request from bs4 import BeautifulSoup with urllib.request.urlopen("") as url: s = url.read() print(s) soup = BeautifulSoup(s, "html.parser") all_tag_a = soup.find_all("a", limit=10) for links in all_tag_a: #print(links.get('href')) print(links) 

One of the possible way to do it:

import urllib ... try: # Python 2 from urllib2 import urlopen except ImportError: # Python 3 from urllib.request import urlopen 

If your code uses Python version 2.x, you can do the following:

from urllib.request import urlopen urlopen(url) 

By the way, I suggest another module called requests, which is more friendly to use. You can use pip install it, and use it like this:

import requests requests.get(url) requests.post(url) 

Use the third-party six module to make your code compatible between Python2 and Python3.

from six.moves import urllib urllib.request.urlopen("<your-url>") 

imgResp = urllib3.request.RequestMethods.urlopen(url) 

Add this RequestMethods before using urlopen