移动Flask-Restplus Swagger API文档-Java 学习之路

我正在尝试使用flask-restplus在python中构建一个restful API . 我希望将swagger文档放在与普通“/”不同的地方 .

我正在按照文档here并按照说明进行操作 . 我正在使用python2.7.3并具有以下代码 ~/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)

当我尝试运行这个 python app.py 时，我得到：

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'

我'm not really sure what exactly is going wrong, I guess I understand that I haven't继承自 Resource ，这通常来自 as_view ，但文档似乎表明这应该有用 .

任何帮助都会得到帮助 .

3 回答

使用Flask-Restplus <= 0.8.0，你应该写：

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)

注意使用 @app 而不是 @api

从v0.8.1（即将发布）开始，你只需要写：

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

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

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

回复于 2024-05-14T16:28:59+08:00

在最近自己挣扎之后，我对这种方法很幸运：

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)

回复于 2024-05-14T16:28:59+08:00

看起来@api需要资源，所以我修改了一些代码来解决错误 . 以下内容仅适用于/ doc /，而不是默认的根级别 .

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)

回复于 2024-05-14T16:28:59+08:00

移动Flask-Restplus Swagger API文档

3 回答

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