如何在 Python 中使用 IDLE 打开当前正在运行的脚本的 .py 文件？

Question

In a GUI app written in Python and Tkinter, I want to add a menu command which will open the source code of the *.py file in IDLE. I also want it to be platform independent.

I've tried using os.system to open IDLE from Scripts folder of Python, but that would be platform dependent. I couldn't find a way to get the Scripts folder of Python to make it independent. How can I achieve this?

Answer 1

To open a file in an IDLE editor, the command line is

<python> -m idlelib <path>

where <python> is the full path to a python executable or a name that resolves to such. If you want to open IDLE with the same python that is running you python app, which is likely what you want, use the platform-independent sys.executable . The path to the source code for the running app is __file__ . This is one of the semi-hidden global variables when python runs a file. Or give the path to any other file.

Whether you use os.system or subprocess is a different issue. The latter is more flexible. subprocess.run replaces os.system . subprocess.Popen should not block. IDLE uses the latter to run user code.

Answer 2

I tried some special way:

    import idlelib.pyshell
    import sys
    sys.argv.clear()
    sys.argv.append("")
    sys.argv.append(fName)
    idlelib.pyshell.main()

It works. But I don't know whether this way will generate some problems.