Details
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Bug
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Status: Closed
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Major
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Resolution: Fixed
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0.18.1, 0.19.0, 0.20.0
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None
Description
What error occurred?
When the Python client times out while reading content from the server, the exception thrown is not "read timeout," but rather "unexpected exception."
Why did this error occur?
the socket.timeout error doesn't includes the errno parameter in its args attribute.
I try to make a timeout error on my pc, it only contains the error string.
Python 3.11.1 (main, Jan 5 2023, 14:25:08) [GCC 11.3.0] Type 'copyright', 'credits' or 'license' for more information IPython 8.13.2 – An enhanced Interactive Python. Type '?' for help.In [1]: import socket ...: from urllib.request import urlopen ...: ...: url = 'http://httpbin.org/get' ...: ...: socket.setdefaulttimeout(0.01) ...: try: ...: urlopen(url) ...: except Exception as e: ...: err1 = e.reason ...: ...:In [2]: type(err1) Out[2]: TimeoutErrorIn [3]: err1.args Out[3]: ('timed out',)
And my system is ubuntu 22.04
ø> uname -a
Linux Macmini 6.2.0-36-generic #37~22.04.1-Ubuntu SMP PREEMPT_DYNAMIC Mon Oct 9 15:34:04 UTC 2 x86_64 x86_64 x86_64 GNU/Linux
TSocket.read handle exception error
Because
socket.error.args == ('timed out',)
, the condition
elif e.args[0] == errno.ETIMEDOUT
cannot be True, resulting in the throwing of the wrong exception
TTransportException(message="unexpected exception", inner=e).
How to fix it
I determine whether a timeout has occurred by checking if the exception is a socket.timeout, no longer relying on e.args.
see details on PR: https://github.com/apache/thrift/pull/2961