Get file name from list of directory contents in Python

Question

I'm connected to Azure Storage and needs to get the name of a file. I have written a function which gets the list of contents in a particular directory.

datepath = np.datetime64('today').astype(str).replace('-', '/')
def list_directory_contents():
    try:
       file_system_client = service_client.get_file_system_client(file_system="container_name")

       paths = file_system_client.get_paths(path = "folder_name/" + str(datepath))

       for path in paths:
           print(path.name + '\n')

    except Exception as e:
     print(e)

And then I'm calling the function

list_directory_contents()

This gives me the something like

folder_name/2020/10/28/file_2020_10_28.csv

Now I want to extract just the file name from the above ie "file_2020_10_28.csv"

Answer 1

You're looking foros.path.basename . It's a cross platform and robust way to do what you want-

>>> os.path.basename("folder_name/2020/10/28/file_2020_10_28.csv")
'file_2020_10_28.csv'

in case it wasn't obvious, the part to be changed in list_directory_contents to print only the basename is this-

for path in paths:
    print(os.path.basename(path.name) + '\n')

assuming path.name is the string that returns something akin to folder_name/2020/10/28/file_2020_10_28.csv