在python脚本中打开文件夹和文件

Question

I have this piece of code:

import os
def find(name):
    for root, dirs, files in os.walk("Desktop/"):
        if name in files:
             os.startfile("Desktop/", name, ".exe")


def findFold(name):
    for root, dirs, files in os.walk("Desktop/"):
        if name in files:
            os.startfile("This PC/", name)

..............
    if OpenFile.lower() == "music" or OpenFile.lower() == "music folder":
        findFold("Music")
..............
    elif OpenFile.lower() == "wolf team" or OpenFile.lower() == "wolfteam":
        find("Wolfteam")

The OpenFile is a string which is the input of the user(the name of the file or folder.
The program doesn't open the files and folders that I want it to open.

I looked in the internet and this is how people said it is and working.. Can somebody help please?

Answer 1

It is very unclear of what you are trying to accomplish, however, I believe you have a backslash where you want a forward slash if you are trying to invoke applications on a Windows based file system

eg I would change

os.startfile("This PC/", name)

to

os.startfile("This PC/{}".format(name))

I also suggest you look into the PEP 8 coding conventions for Python, and use their suggestions for readable code. Namely:

look into adding docstrings to your methods

def fileFold(name):
    '''
    why this method exists
    '''

do not hard code your paths but have them be variables

def findFold(name, prefix="\Desktop\", suffix=".exe")

wrap your activities in Try Except blocks

try:
    os.startfile("{}{}{}".format(prefix, name, suffix))
except OSError as err:
    print("Unable to invoke application {}: {}".format(name, err))