如何创建执行类似以下操作的 api 调用:
api.add_resource(MyResource, '/myresource/<myurlparameter>')
不清楚我如何从内部处理它:
class MyResource(restful.Resource):
def get(self):
print myurlparameter
return ""
此外,我注意到我只能将 add_resource 添加到一个级别:
api.add_resource(MyResource, '/myresource') # this works
api.add_resource(MyResource, '/myresource/test') # this this not work
最佳答案
您可以在 docs 中找到您需要的一切.
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, '/<string:todo_id>')
api.add_resource(HelloWorld,
'/',
'/hello')
关于python - 如何在python中为flask-restful添加参数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21106965/