如何获取在Python中捕获的异常的名称?

问题:如何获取在Python中捕获的异常的名称?

如何获得在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,我捕获了多个(或所有)异常,并想在错误消息中打印异常的名称。

How can I get the name of an exception that was raised in Python?

e.g.,

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

For example, I am catching multiple (or all) exceptions, and want to print the name of the exception in an error message.


回答 0

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'

Here are a few different ways to get the name of the class of the exception:

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

e.g.,

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

回答 1

这行得通,但似乎必须有一种更简单,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'

This works, but it seems like there must be an easier, more direct way?

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

回答 2

您也可以使用sys.exc_info()exc_info()返回3个值:类型,值,回溯。关于文档: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__

You can also use sys.exc_info(). exc_info() returns 3 values: type, value, traceback. On documentation: 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__

回答 3

如果您想要完全限定的类名(例如,sqlalchemy.exc.IntegrityError而不是仅使用IntegrityError),则可以使用下面的函数,该函数是我从MB对另一个问题的出色回答(我只是重命名了一些变量以适合自己的口味)而来:

def get_full_class_name(obj):
    module = obj.__class__.__module__
    if module is None or module == str.__class__.__module__:
        return obj.__class__.__name__
    return module + '.' + obj.__class__.__name__

例:

try:
    # <do something with sqlalchemy that angers the database>
except sqlalchemy.exc.SQLAlchemyError as e:
    print(get_full_class_name(e))

# sqlalchemy.exc.IntegrityError

If you want the fully qualified class name (e.g. sqlalchemy.exc.IntegrityError instead of just IntegrityError), you can use the function below, which I took from MB’s awesome answer to another question (I just renamed some variables to suit my tastes):

def get_full_class_name(obj):
    module = obj.__class__.__module__
    if module is None or module == str.__class__.__module__:
        return obj.__class__.__name__
    return module + '.' + obj.__class__.__name__

Example:

try:
    # <do something with sqlalchemy that angers the database>
except sqlalchemy.exc.SQLAlchemyError as e:
    print(get_full_class_name(e))

# sqlalchemy.exc.IntegrityError

回答 4

此处的其他答案非常适合用于探索目的,但是如果主要目标是记录异常(包括异常的名称),则可以考虑使用logging.exception而不是print?

The other answers here are great for exploration purposes, but if the primary goal is to log the exception (including the name of the exception), perhaps consider using logging.exception instead of print?