AttributeError:类主页没有属性“as_view”
AttributeError:类主页没有属性“as_view”
提问于 2018-10-19 07:42:57
我使用的是Flask Restful，我想在根端点上呈现一个HTML file：
from flask_restful import Resource, Api
app = Flask( __name__, static_url_path = '', static_folder = "docs" )
api = Api(app, catch_all_404s=True)
class Home():
def get(self):
return app.send_static_file( 'index.html' )
api.add_resource(Home, '/')当我运行这段代码时，我会得到以下错误：
api.add_resource(Home, '/')
File "/home/vagrant/.local/lib/python2.7/site-packages/flask_restful/__init__.py", line 404, in add_resource
self._register_view(self.app, resource, *urls, **kwargs)
File "/home/vagrant/.local/lib/python2.7/site-packages/flask_restful/__init__.py", line 444, in _register_view
resource_func = self.output(resource.as_view(endpoint, *resource_class_args,
AttributeError: class Home has no attribute 'as_view'如何在Flask Restful中呈现静态文件？
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-10-19 07:45:34
查看一下烧瓶Restful文档；例如，您可以在文档中找到以下内容：
from flask import Flask, request
from flask_restful import Resource, Api
app = Flask(__name__)
api = Api(app)
todos = {}
class TodoSimple(Resource):
def get(self, todo_id):
return {todo_id: todos[todo_id]}
def put(self, todo_id):
todos[todo_id] = request.form['data']
return {todo_id: todos[todo_id]}
api.add_resource(TodoSimple, '/<string:todo_id>')
if __name__ == '__main__':
app.run(debug=True)正如您在上面看到的，您肯定缺少的一件事是来自Resource的继承。
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原文链接：
https://stackoverflow.com/questions/52887871
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