python - 尝试/使用Python请求模块的正确方法？

Question

try:
    r = requests.get(url, params={'s': thing})
except requests.ConnectionError, e:
    print e #should I also sys.exit(1) after this?

这个对吗？有没有更好的方法来构造它？这会覆盖我所有的基地吗？

Answer 1

看一下Requests exception docs。简而言之:

In the event of a network problem (e.g. DNS failure, refused connection, etc), Requests will raise a ConnectionError exception.

In the event of the rare invalid HTTP response, Requests will raise an HTTPError exception.

If a request times out, a Timeout exception is raised.

If a request exceeds the configured number of maximum redirections, a TooManyRedirects exception is raised.

All exceptions that Requests explicitly raises inherit from requests.exceptions.RequestException.

要回答您的问题，您显示的内容不会涵盖所有基础。您只会捕获与连接有关的错误，而不会发现超时的错误。
捕获异常时该做什么实际上取决于脚本/程序的设计。可以接受退出吗？您可以继续重试吗？如果错误是灾难性的并且您无法继续，那么可以，您可以通过引发 SystemExit(打印错误并调用 sys.exit的一种不错的方法)来中止程序。
您可以捕获基类异常，该异常将处理所有情况:

try:
    r = requests.get(url, params={'s': thing})
except requests.exceptions.RequestException as e:  # This is the correct syntax
    raise SystemExit(e)

或者，您可以分别捕获它们并执行不同的操作。

try:
    r = requests.get(url, params={'s': thing})
except requests.exceptions.Timeout:
    # Maybe set up for a retry, or continue in a retry loop
except requests.exceptions.TooManyRedirects:
    # Tell the user their URL was bad and try a different one
except requests.exceptions.RequestException as e:
    # catastrophic error. bail.
    raise SystemExit(e)

正如 Christian指出的:

If you want http errors (e.g. 401 Unauthorized) to raise exceptions, you can call Response.raise_for_status. That will raise an HTTPError, if the response was an http error.

一个例子:

try:
    r = requests.get('http://www.google.com/nothere')
    r.raise_for_status()
except requests.exceptions.HTTPError as err:
    raise SystemExit(err)

将打印:

404 Client Error: Not Found for url: http://www.google.com/nothere