Hi all:
I wrote the sources below:
from django.views.debug import ExceptionReporter
def recordErrorPage(R, **kw):
if kw.has_key('exc_info'):
reporter = ExceptionReporter(R, *kw['exc_info'])
bug_html = reporter.get_traceback_html()
# bug_html is the content of default Django debug info
page.
else:
bug_html = 'no exc_info'
bp = BugPage(html=bug_html)
bp.save()
t = get_template('customize_500.html')
# this customize_500.html has few information for client.
html = t.render(RequestContext(R, {'bug_page': bp}))
return HttpResponseServerError(html, mimetype='text/html')
handler500 = 'recordErrorPage'
But this method must add a line("param_dict['exc_info'] = exc_info")
in the Django framework library(django/core/handlers/base.py near 166
line)
to let recordErrorPage gets the "exc_info" variable. This is so
dirty.
Perhaps i should use the middleware framework to implement this
feature.
but i read the Django document, the "process_exception" function can
not get the "exc_info" variable, so i can not render a debug info page
like django default page.
Any suggestion?
Thank you so much.
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