Python Urllib Urlopen Not Working

Question

I am just trying to fetch data from a live web by using the urllib module, so I wrote a simple example

Here is my code:

import urllib

sock = urllib.request.urlopen("") 
htmlSource = sock.read()                            
sock.close()                                        
print (htmlSource)

But I got error like:

Traceback (most recent call last):
  File "D:\test.py", line 3, in <module>
    sock = urllib.request.urlopen("") 
AttributeError: 'module' object has no attribute 'request'

Accepted Answer

You are reading the wrong documentation or the wrong Python interpreter version. You tried to use the Python 3 library in Python 2.

Use:

import urllib2

sock = urllib2.urlopen("") 
htmlSource = sock.read()                            
sock.close()                                        
print htmlSource

The Python 2 urllib2 library was replaced by urllib.request in Python 3.

Answer 2

import requests
import urllib

link = ""

f = urllib.request.urlopen(link)
myfile = f.read()

writeFileObj = open('output.xml', 'wb')
writeFileObj.write(myfile)
writeFileObj.close()

Answer 3

In Python3 you can use urllib or urllib3

urllib:

import urllib.request
with urllib.request.urlopen(') as response:
    htmlSource = response.read()

urllib3:

import urllib3
http = urllib3.PoolManager()
r = http.request('GET', ')
htmlSource = r.data

More details could be found in the urllib or python documentation.

Answer 4

This is what i use to get data from urls, its nice because you could save the file at the same time if you need it:

import urllib

result = urllib.urlretrieve("")

print open(result[0]).read()

output:

'<!DOCTYPE html><body style="padding:0; margin:0;"><iframe src="" style="visibility: visible;height: 2000px;" allowtransparency="true" marginheight="0" marginwidth="0" frameborder="0" scrolling="no" width="100%"></iframe></body></html>'

Edit: urlretrieve works in python 2 and 3

Answer 5

Make sure you import requests from urllib, then try this format, it worked for me:

from urllib import request
urllib.request.urlopen( )

Answer 6

I just queried the same question which is now over 5 years old.

Please note that the URL given is also old, so i substituted the python welcome page.

We can use the requests module in python 3.

I use python 3 and the solution is below:

import requests

r = requests.get(')
t = r.text

print(t)

This works and is clean.

Answer 7

For python 3 the correct way should be:

import cv2
import numpy as np
import urllib.request

req = urllib.request.urlopen(')
arr = np.asarray(bytearray(req.read()), dtype=np.uint8)
img = cv2.imdecode(arr, -1) # 'Load it as it is'

cv2.imshow('image_name', img)
if cv2.waitKey() & 0xff == 27: quit()

Here you can find the documentation related to urllib.request

Answer 8

Use this

    import cv2
    import  numpy as np
    import urllib //import urllib using pip
    import requests // import requests using pip`enter code here`
    url = "write your url"
    while True:
    imgresp = urllib.request.urlopen(url)
    imgnp = np.array(bytearray(imgresp.read()),dtype=np.uint8)
    img = cv2.imdecode(imgnp,-1)
    cv2.imshow("test",img)
    cv2.waitKey('q')

Python Urllib Urlopen Not Working

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Python Urllib Urlopen Not Working

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