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Error generating Swagger server (Python Flask) from Swagger editor

2016-05-24 06:24 489 查看

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http://stackoverflow.com/questions/36416679/error-generating-swagger-server-python-flask-from-swagger-editor
I've used the Swagger Editor to manually generate my Swagger spec file and generated the files for a Python Flask server. Following the README I installed connexion, but when I run

python app.py

I get the error: ValueError: need more than 1 value to unpack. Any ideas?

Full stack trace below:

No handlers could be found for logger "connexion.api"
Traceback (most recent call last):
File "app.py", line 5, in <module>
app.add_api('swagger.yaml')
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/app.py", line 144, in add_api
debug=self.debug)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/api.py", line 127, in __init__
self.add_paths()
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/api.py", line 198, in add_paths
six.reraise(*sys.exc_info())
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/api.py", line 187, in add_paths
self.add_operation(method, path, endpoint, path_parameters)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/api.py", line 160, in add_operation
resolver=self.resolver)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/operation.py", line 168, in __init__
resolution = resolver.resolve(self)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/resolver.py", line 50, in resolve
return Resolution(self.resolve_function_from_operation_id(operation_id), operation_id)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/resolver.py", line 71, in resolve_function_from_operation_id
return self.function_resolver(operation_id)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/connexion/utils.py", line 106, in get_function_from_name
module_name, attr_path1 = module_name.rsplit('.', 1)
ValueError: need more than 1 value to unpack

python flask swagger

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asked Apr 5 at 3:36

tw1742
117211

Please open a ticket via github.com/swagger-api/swagger-codegen/issues for tracking. – wing328 Apr 6 at 4:45

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1 Answer 1

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I ran into this as well. From what I see, the generated code from Swagger seems to assume you're using Python 3. While connexion supports both Python 2.7 & 3.4+, it does need a

__init__.py

file in the generated

python-flask-server/

base directory as well as inside the

controllers/

subdirectory to work for Python 2.7 (Implicit Namespace Packages were introduced in Python 3.3). If you create those 2 empty files after generating the code, things should work. If the Swagger generator wants to support Python 2.7 (since connexion allows for it), it would just need to provide those files as well.

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