I have the following structure in my project:
project
├── core
│ └── factory.py
└── docs
└── swagger.yaml
My core.factory.py loads the swagger.yaml in the build_app function with:
setup_swagger(
app,
swagger_url='/docs',
swagger_from_file="/../docs/swagger.yaml"
)
But it crashes the app raising the error gunicorn.errors.HaltServer: <HaltServer 'Worker failed to boot.' 3> (If I remove setup_swagger it works normally).
I have tried:
swagger_from_file="/../docs/swagger.yaml"
swagger_from_file="../docs/swagger.yaml"
swagger_from_file="docs/swagger.yaml"
And they all have failed. So I came to ask for help on that. Is there a way to load the swagger file when that file is located the parent dir?
I have the following structure in my project:
My
core.factory.pyloads the swagger.yaml in thebuild_appfunction with:But it crashes the app raising the error
gunicorn.errors.HaltServer: <HaltServer 'Worker failed to boot.' 3>(If I remove setup_swagger it works normally).I have tried:
swagger_from_file="/../docs/swagger.yaml"swagger_from_file="../docs/swagger.yaml"swagger_from_file="docs/swagger.yaml"And they all have failed. So I came to ask for help on that. Is there a way to load the swagger file when that file is located the parent dir?