Python 3 urllib HTTP Error 412: Precondition Failed - html

Question

I'm trying to parse HTML data of a website. I wrote this code:
import urllib.request
def parse(url):
response = urllib.request.urlopen(url)
html = response.read()
strHTML = html.decode()
return strHTML
website = "http://www.manarat.ac.bd/"
string = parse(website)
but it is showing this error:
Traceback (most recent call last):
File "C:\Users\pupewekate\Videos\RAW\2.py", line 11, in
string = parse(website)
File "C:\Users\pupewekate\Videos\RAW\2.py", line 5, in parse
response = urllib.request.urlopen(url)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 223, in urlopen return opener.open(url, data, timeout)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 532, in open response = meth(req, response)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 642, in http_response 'http', request, response, code, msg,
hdrs)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 570, in error return > self._call_chain(*args)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 504, in _call_chain result = func(*args)
File
"C:\Users\pupewekate\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py",
line 650, in http_error_default raise HTTPError(req.full_url, code,
msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 412: Precondition
Failed
Any solution?

Answer 1

This website checks the user agent header. If it doesn't recognize its value it returns status code 412:
import requests
print(requests.get('http://www.manarat.ac.bd/'))
# <Response [412]>
print(requests.get('http://www.manarat.ac.bd/', headers={'User-Agent': 'Chrome'}))
# <Response [200]>
See this answer for how to set user agent in urlib.

Answer 2

You could use requests module as it is easier to implement, else if you are determined to use urllib, you can use this:
import urllib
def parse(url):
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 6.3;Win64;x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/59.0.3071.115 Safari/537.36'}
response = urllib.request.urlopen(url,headers=headers)
print response
website = "http://www.manarat.ac.bd/"
string = parse(website)