How to open file using argparse?

Question

I want to open file for reading using argparse. In cmd it must look like: my_program.py /filepath

That\'s my try:

parser = argparse.ArgumentParser()
         


        

Answer 1

Take a look at the documentation: https://docs.python.org/3/library/argparse.html#type

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()

print(args.file.readlines())

Answer 2

In order to have the file closed gracefully, you can combine argparse.FileType with the "with" statement

# ....

parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()

with args.file as file:
    print file.read()

--- update ---

Oh, @Wernight already said that in comments

Answer 3

This implementation allows the "file name" parameter to be optional, as well as giving a short description if and when the user enters the -h or --help argument.

parser = argparse.ArgumentParser(description='Foo is a program that does things')
parser.add_argument('filename', nargs='?')
args = parser.parse_args()

if args.filename is not None:
    print('The file name is {}'.format(args.filename))
else:
    print('Oh well ; No args, no problems')

Answer 4

I'll just add the option to use pathlib:

import argparse, pathlib

parser = argparse.ArgumentParser()
parser.add_argument('file', type=pathlib.Path)
args = parser.parse_args()

with args.file.open('r') as file:
    print(file.read())

Answer 5

The type of the argument should be string (which is default anyway). So make it like this:

parser = argparse.ArgumentParser()
parser.add_argument('filename')
args = parser.parse_args()
with open(args.filename) as file:
  # do stuff here