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

PythonPython 3.xUrllib

Python Problem Overview


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("http://www.python.org")
AttributeError: 'module' object has no attribute 'urlopen'

I'm following the guide here: http://www.boddie.org.uk/python/HTML.html

import urllib

file = urllib.urlopen("http://www.python.org")
s = file.read()
f.close()

#I'm guessing this would output the html source code?
print(s)

I'm using Python 3.

Python Solutions


Solution 1 - Python

This works in Python 2.x.

For Python 3 look in the docs:

import urllib.request

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

Solution 2 - Python

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("http://www.python.org") as url:
    s = url.read()

print(s)

Solution 3 - Python

import urllib.request as ur
s = ur.urlopen("http://www.google.com")
sl = s.read()
print(sl)

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

Solution 4 - Python

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 :)

Solution 5 - Python

Change TWO lines:

import urllib.request #line1

#Replace
urllib.urlopen("http://www.python.org")
#To
urllib.request.urlopen("http://www.python.org") #line2

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

siteurl = "http://www.python.org"

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.

Solution 6 - Python

import urllib.request as ur

filehandler = ur.urlopen ('http://www.google.com')
for line in filehandler:
    print(line.strip())

Solution 7 - Python

For python 3, try something like this:

import urllib.request
urllib.request.urlretrieve('http://crcv.ucf.edu/THUMOS14/UCF101/UCF101/v_YoYo_g19_c02.avi', "video_name.avi")

It will download the video to the current working directory

I got help from HERE

Solution 8 - Python

Solution for python3:

from urllib.request import urlopen

url = 'http://www.python.org'
file = urlopen(url)
html = file.read()
print(html)

Solution 9 - Python

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)

Solution 10 - Python

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

Solution 11 - Python

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>")

Solution 12 - Python

import urllib
import urllib.request
from bs4 import BeautifulSoup


with urllib.request.urlopen("http://www.newegg.com/") 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)

Solution 13 - Python

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

Add this RequestMethods before using urlopen

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiondeleteView Question on Stackoverflow
Solution 1 - PythoneumiroView Answer on Stackoverflow
Solution 2 - PythonMartin ThomaView Answer on Stackoverflow
Solution 3 - PythonManu MariarajView Answer on Stackoverflow
Solution 4 - PythonSteven B. PeutzView Answer on Stackoverflow
Solution 5 - PythonShahzaib ChadharView Answer on Stackoverflow
Solution 6 - PythonKamranView Answer on Stackoverflow
Solution 7 - PythonrocksyneView Answer on Stackoverflow
Solution 8 - PythonBanjaliView Answer on Stackoverflow
Solution 9 - Pythonjason.luView Answer on Stackoverflow
Solution 10 - PythonVasyl LyashkevychView Answer on Stackoverflow
Solution 11 - PythonRajat ShuklaView Answer on Stackoverflow
Solution 12 - Pythonuser11649630View Answer on Stackoverflow
Solution 13 - Pythonkhadersha shaikView Answer on Stackoverflow