Flask Restful添加资源参数 [英] Flask Restful add resource parameters

问题描述

我希望将对象实例作为参数传递给 Flask-RESTfull 资源.

I am looking to pass an object instance as a parameter into a Flask-RESTfull Resource.

这是我的设置:

# in main.py
from flask import Flask
from flask.ext.restful import Api
from bar import Bar
from foo import views

app = Flask(__name__)
api = Api(app)

my_bar = Bar()

api.add_resource(views.ApiPage, "/api/my/end/point/")

然后在views.py中，我的资源设置如下:

Then in views.py I have the resource set up as follows:

# In views.py
from flask.ext.restful import Resource

class ApiPage(Resource):
    def get(self):
        serialized = str(my_bar)
        return serialized

所以我遇到的问题是我需要将Bar()的实例传递到api资源中.有没有办法像api.add_resource(views.ApiPage, "/api/my/end/point/", instance=bar)那样通过add_resource方法传递它?

So the issue that I am having is that I need to pass my instance of Bar() into the api resource. Is there any way to pass it in through the add_resource method like api.add_resource(views.ApiPage, "/api/my/end/point/", instance=bar)?

推荐答案

自版本0.3.3(2015年5月22日发布)以来，add_resource()可以将参数传递给您的Resource构造函数.

Since version 0.3.3 (released May 22, 2015), add_resource() is able to pass parameters to your Resource constructor.

在原始示例之后，这是views.py:

Following the original example, here is the views.py:

from flask.ext.restful import Resource

class ApiPage(Resource):
    def __init__(self, bar):
        self.bar = bar

    def get(self):
        serialized = str(my_bar)
        return serialized

以及main.py的相关代码:

# ...
my_bar = Bar()
api.add_resource(views.ApiPage, '/api/my/end/point/',
                 resource_class_kwargs={'bar': my_bar})

这篇关于Flask Restful添加资源参数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IT屋！