What is the appropriate way to intercept WSGI start_response?

Question

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. Curre

Answer 1

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