如何获取Python捕获的异常的名称？

Question

如何获取Python中引发的异常的名称

例如：

try:
    foo = bar
except Exception as exception:
    name_of_exception = ???
    assert name_of_exception == 'NameError'
    print "Failed with exception [%s]" % name_of_exception

例如，我正在捕获多个（或所有）异常，并希望在错误消息中打印该异常的名称

Answer 1

这是可行的，但似乎必须有更简单、更直接的方法

try:
    foo = bar
except Exception as exception:
    assert repr(exception) == '''NameError("name 'bar' is not defined",)'''
    name = repr(exception).split('(')[0]
    assert name == 'NameError'

Answer 2

您也可以使用sys.exc_info()exc_info()返回3个值：类型、值、回溯。关于文件：https://docs.python.org/3/library/sys.html#sys.exc_info

import sys

try:
    foo = bar
except Exception:
    exc_type, value, traceback = sys.exc_info()
    assert exc_type.__name__ == 'NameError'
    print "Failed with exception [%s]" % exc_type.__name__

Answer 3

以下是获取异常类名称的几种不同方法：

1. type(exception).__name__
2. exception.__class__.__name__
3. exception.__class__.__qualname__

例如：

try:
    foo = bar
except Exception as exception:
    assert type(exception).__name__ == 'NameError'
    assert exception.__class__.__name__ == 'NameError'
    assert exception.__class__.__qualname__ == 'NameError'