python - 运行 `python -m unittest` 会更改在回溯中打印覆盖 `__name__` 的异常的方式

Question

我有一段动态创建异常的代码。以这种方式创建的每个异常类都会覆盖其 __name__:

def exception_injector(name, parent, module_dict):
    class product_exception(parent):
        pass
    product_exception.__name__ = name
    product_exception.__module__ = module_dict["__name__"]
    module_dict[name] = product_exception

当我经常使用它时，它可以很好地打印所有内容:

>>> namespace = {"__name__": "some.module"}
>>> exception_injector("TestError", Exception, namespace)
>>> raise namespace["TestError"]("What's going on?")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
some.module.TestError: What's going on?

但是当我使用 unittest 模块时，会打印原始的 __name__:

>>> import unittest
>>> class DemoTestCase(unittest.TestCase):
...     def test_raise(self):
...         namespace = {"__name__": "some.module"}
...         exception_injector("TestError", Exception, namespace)
...         namespace["TestError"]("What's going on?")
...
>>> unittest.main(defaultTest="DemoTestCase", argv=["demo"], exit=False)
E
======================================================================
ERROR: test_raise (__main__.DemoTestCase)
----------------------------------------------------------------------
Traceback (most recent call last):
  File "<stdin>", line 3, in test_raise
some.module.exception_injector.<locals>.product_exception: What's going on?

----------------------------------------------------------------------
Ran 1 test in 0.000s

FAILED (errors=1)
<unittest.main.TestProgram object at 0x1017c3eb8>

unittest 模块可以从异常对象的何处获取原始信息？

Answer 1

单元测试使用 traceback module格式化异常。您可以通过执行相同的操作来重现输出:

>>> import traceback
>>> try:
...     raise namespace["TestError"]("What's going on?")
... except Exception as e:
...     tb = traceback.TracebackException.from_exception(e)
...     print(*tb.format())
...
Traceback (most recent call last):
   File "<stdin>", line 2, in <module>
 some.module.exception_injector.<locals>.product_exception: What's going on?

正在打印的是对象__module__属性值^*，其中object.__qualname__ attribute附加(带有点分隔符)，而不是 __module__ 加 __name__:

>>> namespace["TestError"].__qualname__
'exception_injector.<locals>.product_exception'

限定名称包括 full scope where the class was created (此处为函数名称，但这也可以包括类名称)。

如果您的目标是向模块的全局命名空间添加异常，则只需将其设置为与 name 相同的值即可:

>>> namespace["TestError"].__qualname__ = "TestError"
>>> try:
...     raise namespace["TestError"]("What's going on?")
... except Exception as e:
...     tb = traceback.TracebackException.from_exception(e)
...     print(*tb.format())
...
Traceback (most recent call last):
   File "<stdin>", line 2, in <module>
 some.module.TestError: What's going on?

或者在您的代码上下文中:

def exception_injector(name, parent, module_dict):
    class product_exception(parent, details=args):
        pass

    product_exception.__name__ = name
    product_exception.__qualname__ = name
    product_exception.__module__ = module_dict["__name__"]
    module_dict[name] = product_exception

---

^* 如果模块名称是 __main__ 或 builtins，则回溯模块将忽略异常模块。