问题描述

我正在尝试使用 flask-restplus 在 python 中构建一个宁静的 API.我想让 swagger 文档位于与普通/"不同的位置.

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 "/".

我正在关注这里的文档并按照说明进行操作.我正在使用 python2.7.3 并有以下代码 ~/dev/test/app.py:

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)

当我尝试运行这个 python app.py 我得到:

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'

我不太确定到底出了什么问题，我想我知道我没有从 Resource 继承，而这正是 as_view 通常来自的地方，但文档似乎表明这应该有效.

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.

我们将不胜感激.

推荐答案

使用 Flask-Restplus <= 0.8.0 你应该这样写:

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)

注意使用 @app 而不是 @api

从 v0.8.1(即将发布)开始，您只需编写:

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/')

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

这篇关于移动 Flask-Restplus Swagger API 文档的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！