如何获取python程序的完整路径，包括程序中的文件名? [英] How to get full path to python program including filename within the program?

问题描述

我有一个 python 程序，我想从程序中获取程序的路径，但包括文件名本身.我的文件名为 PyWrapper.py.现在我正在这样做:

I have a python program, and I want to get the path to the program from within the program, but INCLUDING the file name itself. The name of my file is PyWrapper.py. Right now I'm doing this:

import sys,os
pathname = os.path.dirname(sys.argv[0])
fullpath = os.path.abspath(pathname)
print fullpath

输出为:

 /home/mrpickles/Desktop/HANSTUFF/securesdk/src/

这是保存我的文件的目录的路径，但我希望它输出:

This is the path to the directory in which my file is saved, but I would like it to output:

/home/mrpickles/Desktop/HANSTUFF/securesk/src/PyWrapper.py/

哪个是路径，包括文件名本身.这可能吗?谢谢.

Which is the path, including the filename itself. Is this possible? Thanks.

推荐答案

看看 __file__.这会为您提供代码实际所在的文件名.

Take a look at __file__. This gives you a filename where the code actually is.

另外，还有一个选择:

import __main__ as main
print(main.__file__)

这为您提供了正在运行的脚本的文件名(类似于您的argv所做的).

This gives you a filename of a script being run (similar to what your argv does).

当代码被另一个脚本导入时，差异就会发挥作用.

The difference comes into play when the code is imported by another script.

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