Dynamic File Name to a shared folder with open command in python

[bowlofred]

Does this work?

Oddly to me, no. I am surprised. Doubled separators as the first element of a path are supposed to be protected, but it looks like Path() destroys them.

>>> Path("//10.0.0.10", "USB_Storage")
WindowsPath('/10.0.0.10/USB_Storage')
>>> Path(r"\\10.0.0.10", "USB_Storage")
WindowsPath('/10.0.0.10/USB_Storage')

That seems like a bug to me (or I'm not setting something somewhere). If I can get past that limitation and construct the WindowsPath object directly, it should work okay. (And I have to turn the generator into a list to get the length)

>>> p = WindowsPath('//10.0.0.10/USB_Storage')
>>> print(len(list(p.glob("*"))))
87