Move Flask-Restplus Swagger API Docs

问题

I'm trying to use flask-restplus to build a restful API in python. I'd like to have the swagger docs located in a different place than the normal "/".

I'm following the documentation here and have followed the instructions. I'm using python2.7.3 and have the following code ~/dev/test/app.py:

from flask import Flask
from flask.ext.restplus import Api, apidoc

app = Flask(__name__)
api = Api(app, ui=False)

@api.route('/doc/', endpoint='doc')
def swagger_ui():
    return apidoc.ui_for(api)

app.register_blueprint(apidoc.apidoc)

When I try to run this python app.py I get:

Traceback (most recent call last):
  File "app.py", line 7 in <module>
    @api.route('/doc/', endpoint='doc')
  File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restplus/api.py", line 191, in wrapper
    self.add_resources(cls, *urls, **kwargs)
  File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restplus/api.py", line 175, in add_resource
    super(Api, self).add_resource(resource, *urls, **kwargs)
  File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restful/__init__.py", line 396, in add_resource
    self._register_view(self.app, resource, *urls, **kwargs)
  File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restful/__init__.py", line 435, in _register_view
    resource_func = self.output(resource.as_view(endpoint, *resource_class_args,
AttributeError: 'function' object has no attribute 'as_view'

I'm not really sure what exactly is going wrong, I guess I understand that I haven't inherited from Resource which is where as_view would normally come from, but the documentation seems to indicate that this should work.

Any help would be apprecaited.

回答1:

With Flask-Restplus <= 0.8.0 you should write:

from flask import Flask
from flask.ext.restplus import Api, apidoc

app = Flask(__name__)
api = Api(app, ui=False)

@app.route('/doc/', endpoint='doc')
def swagger_ui():
    return apidoc.ui_for(api)

Note the use of a @app instead of @api

Starting from v0.8.1 (soon to be released), you will simply have to write:

from flask import Flask
from flask.ext.restplus import Api, apidoc

app = Flask(__name__)
api = Api(app, doc='/doc/')

See: http://flask-restplus.readthedocs.org/en/latest/swagger.html#swagger-ui

回答2:

After recently struggling with this myself, I've had luck with this approach:

from flask import Flask, Blueprint
from flask.ext.restplus import Api, apidoc

app = Flask(__name__)

blueprint = Blueprint('api', __name__)
api = Api(blueprint, ui=False)

@blueprint.route('/doc/', endpoint='doc')
def swagger_ui():
   return apidoc.ui_for(api)

app.register_blueprint(blueprint)

回答3:

It looks like @api will need Resource, so I modified the code a bit to get around the error. The following will work only at /doc/, not the default root level.

from flask import Flask, make_response
from flask.ext.restplus import Api, apidoc, Resource

app = Flask(__name__)
api = Api(app, ui=False)

@api.route('/doc/', endpoint='doc', doc=False)
class ApiDoc(Resource):
    def get(self):
        return make_response(apidoc.ui_for(api))

app.register_blueprint(apidoc.apidoc)

if __name__ == '__main__':
    app.run(debug=True)

来源：https://stackoverflow.com/questions/31570096/move-flask-restplus-swagger-api-docs

标签

python

Flask

swagger

swagger-ui

flask-restplus