Question

我使用python3和requests包进行了简单的长轮询。它目前看起来像：

def longpoll():
    session = requests.Session()
    while True:
        try:
            fetched = session.get(MyURL)
            input = base64.b64decode(fetched.content)
            output = process(data) 
            session.put(MyURL, data=base64.b64encode(response))
        except Exception as e:
            print(e)
            time.sleep(10)

有一种情况，我想提出一个http错误，而不是process输入和put结果。有没有一种简单的方法可以从高级Session界面执行此操作？或者我是否需要向下钻取以使用较低级别的对象？

Answer 1

由于您可以控制服务器，因此您可能想要撤消第二次呼叫

这是一个使用瓶子来回顾第二轮询的例子

def longpoll():
    session = requests.Session()
    while True: #I'm guessing that the server does not care that we call him a lot of times ...
        try: 
            session.post(MyURL, {"ip_address": my_ip_address}) # request work or I'm alive
            #input = base64.b64decode(fetched.content)
            #output = process(data) 
            #session.put(MyURL, data=base64.b64encode(response))
        except Exception as e:
            print(e)
            time.sleep(10)

@bottle.post("/process") 
def process_new_work():
    data = bottle.request.json()
    output = process(data) #if an error is thrown an HTTP error will be returned by the framework
    return output

这样服务器将获得输出或错误的HTTP状态

从python3请求生成http错误

1 个答案: