如何使用 httplib (python 2.6) 处理超时？

发布于 2024-09-30 11:23:15 字数 414 浏览 5 评论 0原文

我正在使用 httplib 通过 https 访问 api，并且需要在 api 关闭时构建异常处理。

这是一个示例连接：

connection = httplib.HTTPSConnection('non-existent-api.com', timeout=1)
connection.request('POST', '/request.api', xml, headers={'Content-Type': 'text/xml'})
response = connection.getresponse()

这应该超时，因此我期望引发异常，并且 response.read() 仅返回一个空字符串。

我如何知道是否超时？更好的是，优雅地处理第 3 方 api 关闭问题的最佳方法是什么？

原文

I'm using httplib to access an api over https and need to build in exception handling in the event that the api is down.

Here's an example connection:

connection = httplib.HTTPSConnection('non-existent-api.com', timeout=1)
connection.request('POST', '/request.api', xml, headers={'Content-Type': 'text/xml'})
response = connection.getresponse()

This should timeout, so I was expecting an exception to be raised, and response.read() just returns an empty string.

How can I know if there was a timeout? Even better, what's the best way to gracefully handle the problem of a 3rd-party api being down?

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如日中天 2024-10-07 11:23:15

更好的是，优雅地处理第 3 方 API 关闭问题的最佳方法是什么？

API 已关闭、API 返回 http 404、500...是什么意思？

或者您的意思是 API 无法访问时？

首先，我认为您在尝试访问 Web 服务之前无法知道它是否已关闭，因此我建议您可以这样做：

import httplib

conn = httplib.HTTPConnection('www.google.com')  # I used here HTTP not HTTPS for simplify
conn.request('HEAD', '/')  # Just send a HTTP HEAD request 
res = conn.getresponse()

if res.status == 200:
   print "ok"
else:
   print "problem : the query returned %s because %s" % (res.status, res.reason)

并且为了检查 API 是否无法访问，我认为您会这样做最好尝试一下：

import httplib
import socket

try:
   # I don't think you need the timeout unless you want to also calculate the response time ...
   conn = httplib.HTTPSConnection('www.google.com') 
   conn.connect()
except (httplib.HTTPException, socket.error) as ex:
   print "Error: %s" % ex

如果你想要更通用的东西，你可以混合使用两种方法，希望这会有所帮助

Even better, what's the best way to gracefully handle the problem of a 3rd-party api being down?

what's mean API is down , API return http 404 , 500 ...

or you mean when the API can't be reachable ?

first of all i don't think you can know if a web service in general is down before trying to access it so i will recommend for first one you can do like this:

import httplib

conn = httplib.HTTPConnection('www.google.com')  # I used here HTTP not HTTPS for simplify
conn.request('HEAD', '/')  # Just send a HTTP HEAD request 
res = conn.getresponse()

if res.status == 200:
   print "ok"
else:
   print "problem : the query returned %s because %s" % (res.status, res.reason)

and for checking if the API is not reachable i think you will be better doing a try catch:

import httplib
import socket

try:
   # I don't think you need the timeout unless you want to also calculate the response time ...
   conn = httplib.HTTPSConnection('www.google.com') 
   conn.connect()
except (httplib.HTTPException, socket.error) as ex:
   print "Error: %s" % ex

You can mix the two ways if you want something more general ,Hope this will help

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昵称有卵用 2024-10-07 11:23:15

urllib 和 httplib 不公开超时。您必须包括套接字并在那里设置超时：

import socket
socket.settimeout(10) # or whatever timeout you want

urllib and httplib don't expose timeout. You have to include socket and set the timeout there:

import socket
socket.settimeout(10) # or whatever timeout you want

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扭转时空 2024-10-07 11:23:15

这是我发现 httplib2 可以正常工作的。发布它，因为它可能仍然对某人有帮助：

    import httplib2, socket

    def check_url(url):
        h = httplib2.Http(timeout=0.1) #100 ms timeout
        try:
            resp = h.request(url, 'HEAD')
        except (httplib2.HttpLib2Error, socket.error) as ex:
            print "Request timed out for ", url
            return False
        return int(resp[0]['status']) < 400

This is what I found to be working correctly with httplib2. Posting it as it might still help someone :

    import httplib2, socket

    def check_url(url):
        h = httplib2.Http(timeout=0.1) #100 ms timeout
        try:
            resp = h.request(url, 'HEAD')
        except (httplib2.HttpLib2Error, socket.error) as ex:
            print "Request timed out for ", url
            return False
        return int(resp[0]['status']) < 400

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