What is the appropriate way to intercept WSGI start_response?

correctness, middleware, python, wsgi

Solution

You can assign the status as an injected field of `local_start` function itself rather than using `status` list. I used something similar, works fine:

class TransactionalMiddlewareInterface(object):
    def __init__(self, application, **config):
        self.application = application
        self.config = config

    def __call__(self, environ, start_response):
        def local_start(stat_str, headers=[]):
            local_start.status = int(stat_str.split(' ')[0])
            return start_response(stat_str, headers)
        try:
            result = self.application(environ, local_start)
        finally:
            if local_start.status and local_start.status > 199:
                pass

Problem

I have WSGI middleware that needs to capture the HTTP status (e.g. `200 OK`) that inner layers of middleware return by calling `start_response`. Currently I'm doing the following, but abusing a list doesn't seem to be the “right” solution to me: ``` class TransactionalMiddlewareInterface(object): def __init__(self, application, **config): self.application = application self.config = config def __call__(self, environ, start_response): status = [] def local_start(stat_str, headers=[]): status.append(int(stat_str.split(' ')[0])) return start_response(stat_str, headers) try: result = self.application(environ, local_start) finally: status = status[0] if status else 0 if status > 199 and status ``` The reason for the list abuse is that I can not assign a new value to the parent namespace from within a wholly contained function.

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