title: Urllib.request.urlopen().read not working
body: <p>last line of code is giving error.</p>
  
  <pre class="lang-py prettyprint-override"><code>import re
  import urllib.request
  url="https://www.google.com/search?q=google&amp;tbm=fin#scso=_GYPEXIHYJs6gtQXFn7i4Aw2:0"
  data=urllib.request.urlopen(url).read()
  </code></pre>
  
  <blockquote>
    <p>url="https://www.google.com/search?q=google&amp;tbm=fin#scso=_GYPEXIHYJs6gtQXFn7i4Aw2:0"
    data=urllib.request.urlopen(url).read()
    Traceback (most recent call last):
      File "", line 1, in 
        data=urllib.request.urlopen(url).read()
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 222, in urlopen
        return opener.open(url, data, timeout)
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 531, in open
        response = meth(req, response)
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 641, in http_response
        'http', request, response, code, msg, hdrs)
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 569, in error
        return self._call_chain(*args)
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 503, in _call_chain
        result = func(*args)
      File "C:\Users\SHARM\AppData\Local\Programs\Python\Python37-32\lib\urllib\request.py", line 649, in http_error_default
        raise HTTPError(req.full_url, code, msg, hdrs, fp)
    urllib.error.HTTPError: HTTP Error 403: Forbidden</p>
  </blockquote>
  
  <p>I want to open the url. Url works when i open manually. I do not understand why i get this error and what does this error mean?</p>