I've just realized, that a bypass is possible to achieve the goal I need
without redefining sys.excepthook:
def actual_application(environ, start_response):
# Real WSGI application goes here
start_response('200 OK', [('Content-Type', 'text/plain')])
return ['OK']
def error(environ, start_response):
# Format a trace and a debug info to send it to a browser here
start_response('200 OK', [('Content-Type', 'text/plain')])
return ['Error']
def application(environ, start_response):
# And the mod_wsgi application hook should consist of the only try...except
block
try:
return actual_application(environ, start_response)
except:
return error(environ, start_response)
But the question why I cannot redefine sys.excepthook successfully is still
unanswered.
On Wednesday, February 18, 2015 at 10:03:48 AM UTC+3, Alexander Sh wrote:
>
>
>
> <http://stackoverflow.com/questions/28570523/handling-exceptions-under-mod-wsgi-assigning-to-sys-excepthook-doesnt-work#>
>
>
> I want to generate an html page with http response status '200 OK' in case
> of an uncaught exception in my wsgi application, instead of web server's
> standard '500 Internal Server Error' response for such case. To do so, I've
> made the following sample code:
>
> def application(environ, start_response):
> def exception(etype, evalue, trace):
> start_response('200 OK', [('Content-Type', 'text/plain')])
> return ['Error']
>
> sys.excepthook = exception
>
> ...
>
>
> But the function 'exception' is never called, and a standard '500 Internal
> Server Error' response is still generated by server in case of an uncaught
> exception.
>
> I looked through the documentation, but unable to find the answer. Are
> there any ways to handle uncaught by try..except exceptions under mod_wsgi?
>
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