在Python 3中打开文本文件

Question

I'm not sure if this has been answered already but I did look and can't find any.

My teammates does not have access to a file via this path: \\\\SYDSFILES01\\Product\\MNL\\SellOutQC\\PythonScripts\\ . They have access to all files within the MNL folder but cannot access \\\\sydsfiles01\\product\\ . What we did is to map the folder ending up with N:\\SellOutQC\\PythonScripts\\ .

However, when I run the below, it results to an error as the program tries to locate the file via \\\\SYDSFILES01\\Product\\MNL\\SellOutQC\\PythonScripts\\ . I tried some workaround but to no avail. I hope you could help. The script being run is within N:\\SellOutQC\\

def import_weekly():
    import csv
    import os
    file = "/PythonScripts/parameters.txt"
    path = os.getcwd()+file

    d={}
    with open(path, 'r+') as file:
        for i in csv.reader(file,delimiter='\t'):
            d[i[0]]=i[1]
    return d

Error: 
PermissionError: [Errno 13] Permission denied: '\\\\sydsfiles01\\product\\manila\\selloutqc\\pythonscripts\\Parameters.txt'

Answer 1

I assume that your code is working with the hardcoded path r'N:\\SellOutQC\\PythonScripts\\Parameters.txt' , you can use os.popen('cd').readline().strip('\\n') to read the mapped location of current working directory on Windows system instead of os.getcwd() . eg

mapped_path = os.popen("cd").readline().strip('\\n')

The print(mapped_path) should print N:\\\\SellOutQC in your case.