urllib.error.URLError: python 3

Question

I have this code in python 3.4

import urllib.request


def read_text():
    document = open(r"C:\mystuff\udacity\pruebatxt.txt")
    lec = document.read()
    document.close()
    print(lec)
    check_profanity(lec)

def check_profanity (text_to_check):
    print(text_to_check)

finally, i found the sollution by changing follow line

    lecture = urllib.request.urlopen("http://www.wdyl.com/profanity?q="+ urllib.parse.quote(text_to_check))
    output=lecture .read()
    print(output)

    lecture.close()
read_text()

And it gives me this error:

  File "C:\Python34\lib\urllib\request.py", line 1159, in do_request_
    raise URLError('no host given')
urllib.error.URLError: <urlopen error no host given>

What could be wrong?

Martijn Pieters · Accepted Answer · 2014-07-22T22:43:20.787

3

urlopen() returns a response, not a request. You cannot pass that to urlopen() again.

Remove the redundant urlopen() call:

lecture = urllib.request.urlopen("http://www.wdyl.com/profanity?q="+text_to_check)
output = lecture.read()

You may have a broken proxy configuration; verify that your proxies are sane by inspecting what Python finds in the Windows registry with:

print(urllib.request.getproxies())

or bypass proxy support altogether with:

lecture = urllib.request.urlopen(
    "http://www.wdyl.com/profanity?q="+text_to_check
    proxies={})

edited Jul 22 '14 at 22:43

answered Jul 22 '14 at 21:40

Martijn Pieters

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What does `print(urllib.request.getproxies())` show? Does it work when you add `proxies={}` to the `urlopen()` call? – Martijn Pieters Jul 22 '14 at 22:40
Thanks for your help, in the code i change a line and i found the solution – Jasher Jul 23 '14 at 21:25

score 1 · Answer 2 · answered Jul 14 '15 at 18:17

Your code returns a binary string. It would be better to decode it into ascii format:

with urllib.request.urlopen("http://www.wdyl.com/profanity?q=" + urllib.parse.quote(text_to_check)) as response:
    output = response.read()
output = output.decode('ascii')
print (output)

urllib.error.URLError: python 3

finally, i found the sollution by changing follow line

2 Answers2