my url is

http://abc.com/search/search.html?search=汽车

i get a exception below:



Traceback (most recent call last):

  File "/usr/lib/python2.5/site-packages/django/core/servers/
basehttp.py", line 277, in run
    self.result = application(self.environ, self.start_response)

  File "/usr/lib/python2.5/site-packages/django/core/servers/
basehttp.py", line 634, in __call__
    return self.application(environ, start_response)

  File "/usr/lib/python2.5/site-packages/django/core/handlers/
wsgi.py", line 226, in __call__
    response = middleware_method(request, response)

  File "/usr/lib/python2.5/site-packages/django_authopenid-0.9.6-
py2.5.egg/django_authopenid/middleware.py", line 19, in
process_response
    path = request.get_full_path()

  File "/usr/lib/python2.5/site-packages/django/core/handlers/
wsgi.py", line 119, in get_full_path
    return '%s%s' % (self.path, self.environ.get('QUERY_STRING', '')
and ('?' + self.environ.get('QUERY_STRING', '')) or '')

UnicodeDecodeError: 'ascii' codec can't decode byte 0xe6 in position
8: ordinal not in range(128)


so i need to encode my non-English charactor.and decode it .




On Sep 3, 3:21 pm, babatu <[EMAIL PROTECTED]> wrote:
> also need to decode it.
>
> On Sep 3, 3:19 pm, babatu <[EMAIL PROTECTED]> wrote:
>
> > How to encode the url ??make the non-English looks like %CD%E2%C6%C5%BC
> > %D2
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