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Convert Unicode to ASCII without errors in Python

My code just scrapes a web page, then converts it to Unicode.

html = urllib.urlopen(link).read()
html.encode("utf8","ignore")
self.response.out.write(html)

But I get a UnicodeDecodeError:


Traceback (most recent call last):
  File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/ext/webapp/__init__.py", line 507, in __call__
    handler.get(*groups)
  File "/Users/greg/clounce/main.py", line 55, in get
    html.encode("utf8","ignore")
UnicodeDecodeError: 'ascii' codec can't decode byte 0xa0 in position 2818: ordinal not in range(128)

I assume that means the HTML contains some wrongly-formed attempt at Unicode somewhere. Can I just drop whatever code bytes are causing the problem instead of getting an error?

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1 Answer

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>>> u'aあ?'.encode('ascii', 'ignore')
'a'

Decode the string you get back, using either the charset in the the appropriate meta tag in the response or in the Content-Type header, then encode.

The method encode(encoding, errors) accepts custom handlers for errors. The default values, besides ignore, are:

>>> u'aあ?'.encode('ascii', 'replace')
b'a??'
>>> u'aあ?'.encode('ascii', 'xmlcharrefreplace')
b'aあä'
>>> u'aあ?'.encode('ascii', 'backslashreplace')
b'a\u3042\xe4'

See https://docs.python.org/3/library/stdtypes.html#str.encode


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