如果名称存在，如何使用“01”创建文件？

Question

Is there a specific argument in open() built-in function so that if the filename already exists, it creates a file by adding a number to its name??

such that if "file.txt" exists, it automatically creates "file-01.txt"

Or any other solution.!

Answer 1

No, I don't think there's something like this but you can do it yourself using os.path.isfile :

import os
filename = "yourFileName.txt"
if os.path.isfile(filename): #check if filename exists in the directory
    filename = filename.split(".")[:-1] + "-01" + filename.split(".")[-1]
with open(filename, "w+") as f:
    f.write(yourString)

Answer 2

Something like this?

import os
if os.path.exists(filename):
   fileparts = filename.split('.')
   filename = fileparts[0] + '01.'
   for a in fileparts[1:]:
     filename += a

Answer 3

I have found a solution, Thanks!!

b = True
c = 1
while b:
    f_name = 'Task-{:02.0f}.txt'.format(c)
    try:
        f = open(f_name,'x')
        b = False
    except FileExistsError:
        c += 1
f.close()

Answer 4

See what you think of this. ...this is what I use to do what you're looking for. It's the smallest way I found to solve the problem before, and is easy to wrap into a function:

import os

name = 'blah.txt'

uniq_name = name
while os.path.isfile(uniq_name):
    # if increment variable 'delta' isn't defined, make it 1.  Otherwise increment
    delta = delta+1 if 'delta' in vars() else 1
    uniq_name = f'{os.path.splitext(name)[0]}-{delta}{os.path.splitext(name)[1]}'

# this you don't need - it's just equivalent to a 'touch' command to show
# the output
open(uniq_name, 'a').close()