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在Flask-Restful中调用资源的2种方法?

[英]2 way to call a resource in Flask-Restful?

I'm learning and using Python, Flask and Flask-Restful for traineeship and I got a question : 我正在学习并使用Python,Flask和Flask-Restful进行实习,但我遇到了一个问题:

Right now I've something like this 现在我有这样的事情

  class CheckTXT(Resource):
  def get(self):

    import dns.resolver

    dmn = request.args.get('dmn')

    answers = dns.resolver.query(dmn, 'TXT')

    c = []

    for rdata in answers:
        for txt_string in rdata.strings:
            c.append(txt_info(dmn, txt_string))

    end = time.time()
    tm = end - start

    return lookup("TXT", dmn, c, tm)

and

api.add_resource(CheckTXT, '/lookup/txt')

I'd like to call it by 2 way : 我想用两种方式称呼它:

lookup/txt?dmn=stackoverflow.com

/lookup/txt/stackoverflow.com

The first one is working but I don't know how to do the second or even if it's possible. 第一个正在工作,但是我不知道如何做第二个,即使可能的话。

Someone can help me ? 有人可以帮助我吗? Thanks for your attention and your patience ! 感谢您的关注和耐心等待! You're helping a young padawan ahah 你在帮一个年轻的padawan啊

Yes you can use below as endpoint and possible with flask resful 是的,您可以在下面用作端点,并可能在烧瓶稳定时使用

/lookup/txt/stackoverflow.com

for that you need to add resources like 为此,您需要添加资源,例如

api.add_resource(CheckTXT, '/lookup/txt/<string:name>') and you can access that field in your implementation like below api.add_resource(CheckTXT, '/lookup/txt/<string:name>') ,您可以像下面这样在实现中访问该字段

class CheckTXT(Resource):
  def get(self,name):
      print name 

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