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Fix typo

* Use the compatible way to handle the exception. You can find the
source code wsgi_app in app.py, and it use the compatible way, so update it
* Fix typo in config.py
* Fix typo in app.py
pull/1716/head
lord63 9 years ago
parent
commit
6d0bbd627c
  1. 2
      docs/reqcontext.rst
  2. 2
      flask/app.py
  3. 2
      flask/config.py

2
docs/reqcontext.rst

@ -69,7 +69,7 @@ find a piece of code that looks very much like this::
with self.request_context(environ):
try:
response = self.full_dispatch_request()
except Exception, e:
except Exception as e:
response = self.make_response(self.handle_exception(e))
return response(environ, start_response)

2
flask/app.py

@ -421,7 +421,7 @@ class Flask(_PackageBoundObject):
#: A dictionary with lists of functions that should be called after
#: each request. The key of the dictionary is the name of the blueprint
#: this function is active for, ``None`` for all requests. This can for
#: example be used to open database connections or getting hold of the
#: example be used to close database connections or getting hold of the
#: currently logged in user. To register a function here, use the
#: :meth:`after_request` decorator.
self.after_request_funcs = {}

2
flask/config.py

@ -222,7 +222,7 @@ class Config(dict):
app.config['IMAGE_STORE_BASE_URL'] = 'http://img.website.com'
image_store_config = app.config.get_namespace('IMAGE_STORE_')
The resulting dictionary `image_store` would look like::
The resulting dictionary `image_store_config` would look like::
{
'type': 'fs',

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