在Python中，如何捕获警告，就像是异常一样？ [英] In Python, how does one catch warnings as if they were exceptions?

问题描述

在我的python代码中使用的第三方库（以C编写）正在发出警告。我想要使​​用 try 除了语法以正确处理这些警告。有没有办法这样做？

A third-party library (written in C) that I use in my python code is issuing warnings. I want to be able to use the try except syntax to properly handle these warnings. Is there a way to do this?

推荐答案

引用python手册（ 27.6.4。测试警告）：

To quote from the python handbook (27.6.4. Testing Warnings):

import warnings

def fxn():
    warnings.warn("deprecated", DeprecationWarning)

with warnings.catch_warnings(record=True) as w:
    # Cause all warnings to always be triggered.
    warnings.simplefilter("always")
    # Trigger a warning.
    fxn()
    # Verify some things
    assert len(w) == 1
    assert issubclass(w[-1].category, DeprecationWarning)
    assert "deprecated" in str(w[-1].message)

（编辑：固定的例子，是一个部分关闭）

(edit: fixed example, was a section off)

这篇关于在Python中，如何捕获警告，就像是异常一样？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！