Python stdin filename

Question

I'm trying to get the filename thats given in the command line. For example:

python3 ritwc.py < DarkAndStormyNight.txt

I'm trying to get DarkAndStormyNight.txt

When I try fileinput.filename() I get back same with sys.stdin. Is this possible? I'm not looking for sys.argv[0] which returns the current script name.

Thanks!

Answer 1

In general it is not possible to obtain the filename in a platform-agnostic way. The other answers cover sensible alternatives like passing the name on the command-line.

On Linux, and some related systems, you can obtain the name of the file through the following trick:

import os
print(os.readlink('/proc/self/fd/0'))

/proc/ is a special filesystem on Linux that gives information about processes on the machine. self means the current running process (the one that opens the file). fd is a directory containing symbolic links for each open file descriptor in the process. 0 is the file descriptor number for stdin.

Answer 2

You can use ArgumentParser, which automattically gives you interface with commandline arguments, and even provides help, etc

from argparse import ArgumentParser
parser = ArgumentParser()                                                                 
parser.add_argument('fname', metavar='FILE', help='file to process')
args = parser.parse_args()

with open(args.fname) as f:
    #do stuff with f

Now you call python2 ritwc.py DarkAndStormyNight.txt. If you call python3 ritwc.py with no argument, it'll give an error saying it expected argument for FILE. You can also now call python3 ritwc.py -h and it will explain that a file to process is required.

PS here's a great intro in how to use it: http://docs.python.org/3.3/howto/argparse.html