Python:如何为子目录中的所有源文件运行 unittest.main()? [英] Python: How to run unittest.main() for all source files in a subdirectory?

问题描述

我正在开发一个包含多个源文件的 Python 模块，每个源文件都有自己的测试类，这些测试类派生自 unittest 就在源代码中.考虑目录结构:

I am developing a Python module with several source files, each with its own test class derived from unittest right in the source. Consider the directory structure:

dirFoo\
    test.py
    dirBar\
        __init__.py
        Foo.py
        Bar.py

要测试 Foo.py 或 Bar.py，我会在 Foo.py 和 Bar.py 源文件的末尾添加:

To test either Foo.py or Bar.py, I would add this at the end of the Foo.py and Bar.py source files:

if __name__ == "__main__":
    unittest.main()

并在任一源上运行 Python，即

And run Python on either source, i.e.

$ python Foo.py
...........
----------------------------------------------------------------------
Ran 11 tests in 2.314s

OK

理想情况下，我会让test.py"自动搜索任何 unittest 派生类的 dirBar 并调用unittest.main()".在实践中做到这一点的最佳方法是什么?

Ideally, I would have "test.py" automagically search dirBar for any unittest derived classes and make one call to "unittest.main()". What's the best way to do this in practice?

我尝试使用 Python 为每个 *.py 文件调用 execfiledirBar，它为找到的第一个 .py 文件运行一次 &退出调用 test.py，然后我必须通过在每个源文件中添加 unittest.main() 来复制我的代码——这违反了 DRY 原则.

I tried using Python to call execfile for every *.py file in dirBar, which runs once for the first .py file found & exits the calling test.py, plus then I have to duplicate my code by adding unittest.main() in every source file--which violates DRY principles.

推荐答案

我知道有一个显而易见的解决方案:

I knew there was an obvious solution:

dirFoo\
    __init__.py
    test.py
    dirBar\
        __init__.py
        Foo.py
        Bar.py

dirFoo/test.py 的内容

Contents of dirFoo/test.py

from dirBar import *
import unittest

if __name__ == "__main__":

    unittest.main()

运行测试:

$ python test.py
...........
----------------------------------------------------------------------
Ran 11 tests in 2.305s

OK

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