Pyinstaller EXE的__file__指向一个.py文件

Question

Situation: My Python script has a line of code that copies itself to another directory

shutil.copyfile(os.path.abspath(__file__), newPath)

Problem: The script is then compiled into an EXE and ran. The error given is as follows:

FileNotFoundError: No such file or Directory: "C:\Path\To\EXE\script.py"

As you can see, the EXE is looking for the .py version of itself (ie uncompiled version)

Question: Is there another Python command that can still let the executable find itself and not the .py version of itself?

Additional information: I was going to try to just replace .py with .exe and see if it works -- it would have if not for the program to fail if I change the name of the executable.

C:\ > script.exe
#Works as expected

C:\ > ren script.exe program.exe
C:\ > program.exe
FileNotFoundError: No such file or directory: "C:\script.py"

Answer 1

I was stuck in this problem too. Finally I found the solution from the official document .

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Solution :

Use sys.argv[0] or sys.executable to access the real path of the executed file.

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Explanation :

This is because your executable is a bundle environment. In this environment, all the __file__ constants are relative paths relative to a virtual directory (actually the directory where the initial entrance file lies in).

As instructed by the document, you can access the absolute by using sys.argv[0] or sys.executable , as they are pointing to the actually executed command. So in a bundle environment, you call script.exe , and sys.executable will be script.exe . While in a running live environment, you call python script.path , and sys.executable will be python.exe .

Answer 2

Try the following:

from os.path import abspath, splitext
fromfile_without_ext, ext = splitext(abspath(__file__))
shutil.copyfile(fromfile_without_ext + '.exe', newPath)

(Did not test it, but should work...)