What is the appropriate way to intercept WSGI start_response?

女生的网名这么多〃 提交于 2019-12-01 05:50:06

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
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