Question

Pytest的断言失败输出比Python中的默认输出更具信息性和有用性。我想在正常运行Python程序时利用此功能，而不仅仅是在执行测试时使用。有没有办法在我的脚本中覆盖Python的assert行为，以便在仍以python script/pytest_assert.py运行程序的同时使用pytest打印堆栈跟踪信息？

示例程序

def test_foo():
  foo = 12
  bar = 42
  assert foo == bar

if __name__ == '__main__':
  test_foo()

$ python script/pytest_assert.py

Traceback (most recent call last):
  File "script/pytest_assert.py", line 8, in <module>
    test_foo()
  File "script/pytest_assert.py", line 4, in test_foo
    assert foo == bar
AssertionError

$ pytest script/pytest_assert.py

======================== test session starts ========================
platform linux -- Python 3.5.3, pytest-3.3.2, py-1.5.2, pluggy-0.6.0
rootdir: /usr/local/google/home/danijar, inifile:
collected 1 item                                                    

script/pytest_assert.py F                                     [100%]

============================= FAILURES ==============================
_____________________________ test_foo ______________________________

    def test_foo():
      foo = 12
      bar = 42
>     assert foo == bar
E     assert 12 == 42

script/pytest_assert.py:4: AssertionError
===================== 1 failed in 0.02 seconds =====================

所需结果

$ python script/pytest_assert.py

Traceback (most recent call last):
  File "script/pytest_assert.py", line 8, in <module>
    test_foo()

    def test_foo():
      foo = 12
      bar = 42
>     assert foo == bar
E     assert 12 == 42

script/pytest_assert.py:4: AssertionError

进度更新

我最接近的是这个，但它仅适用于该函数中的断言并向跟踪发送垃圾邮件：

import ast
import inspect

from _pytest import assertion


def test_foo():
  foo = []
  foo.append(13)
  foo = foo[-1]
  bar = 42
  assert foo == bar, 'message'


if __name__ == '__main__':
  tree = ast.parse(inspect.getsource(test_foo))
  assertion.rewrite.rewrite_asserts(tree)
  code = compile(tree, '<name>', 'exec')
  ns = {}
  exec(code, ns)
  ns[test_foo.__name__]()

$ python script/pytest_assert.py
Traceback (most recent call last):
  File "script/pytest_assert.py", line 21, in <module>
    ns[test_foo.__name__]()
  File "<name>", line 6, in test_foo
AssertionError: message
assert 13 == 42

Answer 1

免责声明

尽管肯定有一种方法可以重用pytest代码以所需格式打印回溯，但是您需要使用的东西不是公共API的一部分，因此生成的解决方案将太脆弱，需要调用不相关的pytest代码（出于初始化目的），可能会导致软件包更新中断。最好的选择是重写pytest代码作为示例。

注释

基本上，下面的概念验证代码可以完成三件事：

将默认的sys.excepthook替换为自定义的ExceptionInfo.getrepr()：这是更改默认回溯格式的必要条件。示例：

import sys

orig_hook = sys.excepthook

def myhook(*args):
    orig_hook(*args)
    print('hello world')

if __name__ == '__main__':
    sys.excepthook = myhook
    raise ValueError()

将输出：

Traceback (most recent call last):
  File "example.py", line 11, in <module>
    raise ValueError()
ValueError
hello world

将打印格式化的异常信息，而不是hello world。为此，我们使用this old article。
要访问断言中的其他信息，pytest会重写assert语句（您可以在PEP 302中重写后获得一些粗略的信息）。为此，pytest注册了Config中指定的自定义导入挂钩。挂钩是最有问题的部分，因为它与AssertionRewriter对象紧密耦合，我也注意到一些模块导入会引起问题（我猜想pytest不会失败，仅是因为模块已经被导入了）当挂钩被注册时；将尝试编写一个测试，该测试在pytest运行时重现该问题并创建一个新的问题）。因此，我建议编写一个调用{{3}}的自定义导入挂钩。这个AST树漫游者类是断言重写中必不可少的部分，而AssertionRewritingHook并不那么重要。

代码

so-51839452
├── hooks.py
├── main.py
└── pytest_assert.py

`hooks.py`

import sys

from pluggy import PluginManager
import _pytest.assertion.rewrite
from _pytest._code.code import ExceptionInfo
from _pytest.config import Config, PytestPluginManager


orig_excepthook = sys.excepthook

def _custom_excepthook(type, value, tb):
    orig_excepthook(type, value, tb)  # this is the original traceback printed
    # preparations for creation of pytest's exception info
    tb = tb.tb_next  # Skip *this* frame
    sys.last_type = type
    sys.last_value = value
    sys.last_traceback = tb

    info = ExceptionInfo(tup=(type, value, tb, ))

    # some of these params are configurable via pytest.ini
    # different params combination generates different output
    # e.g. style can be one of long|short|no|native
    params = {'funcargs': True, 'abspath': False, 'showlocals': False,
              'style': 'long', 'tbfilter': False, 'truncate_locals': True}
    print('------------------------------------')
    print(info.getrepr(**params))  # this is the exception info formatted
    del type, value, tb  # get rid of these in this frame


def _install_excepthook():
    sys.excepthook = _custom_excepthook


def _install_pytest_assertion_rewrite():
    # create minimal config stub so AssertionRewritingHook is happy
    pluginmanager = PytestPluginManager()
    config = Config(pluginmanager)
    config._parser._inidict['python_files'] = ('', '', [''])
    config._inicache = {'python_files': None, 'python_functions': None}
    config.inicfg = {}

    # these modules _have_ to be imported, or AssertionRewritingHook will complain
    import py._builtin
    import py._path.local
    import py._io.saferepr

    # call hook registration
    _pytest.assertion.install_importhook(config)

# convenience function
def install_hooks():
    _install_excepthook()
    _install_pytest_assertion_rewrite()

`main.py`

调用hooks.install_hooks()之后，main.py将修改回溯打印。在install_hooks()调用之后导入的每个模块都将在导入时重写断言。

from hooks import install_hooks

install_hooks()

import pytest_assert


if __name__ == '__main__':
    pytest_assert.test_foo()

`pytest_assert.py`

def test_foo():
    foo = 12
    bar = 42
    assert foo == bar

示例输出

$ python main.py
Traceback (most recent call last):
  File "main.py", line 9, in <module>
    pytest_assert.test_foo()
  File "/Users/hoefling/projects/private/stackoverflow/so-51839452/pytest_assert.py", line 4, in test_foo
    assert foo == bar
AssertionError
------------------------------------
def test_foo():
        foo = 12
        bar = 42
>       assert foo == bar
E       AssertionError

pytest_assert.py:4: AssertionError

总结

我会写自己的AssertionRewritingHook版本，而没有整个无关的pytest东西。 AssertionRewriter看起来非常可重用；尽管它需要一个Config实例，但是它仅用于警告打印，可以保留给None。

有了这些，编写自己的函数以正确格式化异常，替换sys.excepthook，就可以完成。

我可以修补Python的断言以获取py.test提供的输出吗？