如何创建一个类似以下内容的api调用：

api.add_resource(MyResource, '/myresource/')

目前尚不清楚如何从内部进行处理：

class MyResource(restful.Resource):

def get(self):

print myurlparameter

return ""

另外,我注意到我只能将add_resource提升到一个级别：

api.add_resource(MyResource, '/myresource') # this works

api.add_resource(MyResource, '/myresource/test') # this this not work

解决方法:

您可以在docs中找到所需的一切.

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, '/')

api.add_resource(HelloWorld,

'/',

'/hello')

标签：flask-restful,flask,python

来源： https://codeday.me/bug/20191122/2058270.html