Python：导入模块而不执行脚本 [英] Python: import module without executing script

问题描述

我查看了类似的问题，但确实如此没有真正回答我的问题。假设我有以下代码（过于简化以仅突出显示我的问题）。

I looked at a similar question but it does not really answer the question that I have. Say I have the following code (overly simplified to highlight only my question).

class A:
    def __init__(self,x):
        self.val = x

a = A(4)
print a.val

此代码位于文件 someones_class.py 中。我现在想在我的程序中导入并使用类 A ，而无需修改 someones_class.py 。如果我从someones_class导入A ，python仍会执行文件中的脚本行。

This code resides in a file someones_class.py. I now want to import and use class A in my program without modifying someones_class.py. If I do from someones_class import A, python would still execute the script lines in the file.

问题：是有没有办法只导入类 A 而不执行最后两行？

Question: Is there a way to just import class A without the last two lines getting executed?

我知道 if __name__ =='__ main __'但是我没有选择修改 someones_class.py 文件，因为只有在获得我的程序开始执行。

I know about if __name__ == '__main__' thing but I do not have the option of modifying someones_class.py file as it is obtained only after my program starts executing.

推荐答案

这个答案只是为了证明可以完成，但是显然需要一个更好的解决方案，以确保你包括你想要包括的类。

This answer is just to demonstrate that it can be done, but would obviously need a better solution to ensure you are including the class(es) you want to include.

>>> code = ast.parse(open("someones_class.py").read())
>>> code.body.pop(1)
<_ast.Assign object at 0x108c82450>
>>> code.body.pop(1)
<_ast.Print object at 0x108c82590>
>>> eval(compile(code, '', 'exec'))
>>> test = A(4)
>>> test
<__main__.A instance at 0x108c7df80>

您可以检查代码正文你想要包含的元素，然后移除剩下的元素。

You could inspect the code body for the elements you want to include and remove the rest.

注意：这是一个巨大的黑客。

NOTE: This is a giant hack.

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