python - unicode().decode('utf-8', 'ignore') raising UnicodeEncodeError -

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here code:

>>> z = u'\u2022'.decode('utf-8', 'ignore') traceback (most recent call last):   file "<stdin>", line 1, in <module>   file "/usr/lib/python2.6/encodings/utf_8.py", line 16, in decode     return codecs.utf_8_decode(input, errors, true) unicodeencodeerror: 'latin-1' codec can't encode character u'\u2022' in position 0: ordinal not in range(256)

why unicodeencodeerror raised when using .decode?

why error raised when using 'ignore'?

when first started messing around python strings , unicode, took me awhile understand jargon of decode , encode too, here's post here may help:

think of decoding go regular bytestring to unicode , encoding from unicode. in other words:

you de - code str produce unicode string

and en - code unicode string produce str.

so: 

unicode_char = u'\xb0'  encodedchar = unicode_char.encode('utf-8')

encodedchar contain unicode character, displayed in selected encoding (in case, utf-8).


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