在Python 2.7中,给定一个URL如
尝试过
example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0
,如何将其解码为期望的结果example.com?title==правовая+защита
?尝试过
url=urllib.unquote(url.encode("utf8"))
, 但似乎给出了错误的结果。