Fix Python – How do I pass a string into subprocess.Popen (using the stdin argument)?


Asked By – Daryl Spitzer

If I do the following:

import subprocess
from cStringIO import StringIO

I get:

Traceback (most recent call last):
  File "<stdin>", line 1, in ?
  File "/build/toolchain/mac32/python-2.4.3/lib/python2.4/", line 533, in __init__
    (p2cread, p2cwrite,
  File "/build/toolchain/mac32/python-2.4.3/lib/python2.4/", line 830, in _get_handles
    p2cread = stdin.fileno()
AttributeError: 'cStringIO.StringI' object has no attribute 'fileno'

Apparently a cStringIO.StringIO object doesn’t quack close enough to a file duck to suit subprocess.Popen. How do I work around this?

Now we will see solution for issue: How do I pass a string into subprocess.Popen (using the stdin argument)?


Popen.communicate() documentation:

Note that if you want to send data to
the process’s stdin, you need to
create the Popen object with
stdin=PIPE. Similarly, to get anything
other than None in the result tuple,
you need to give stdout=PIPE and/or
stderr=PIPE too.

Replacing os.popen*

    pipe = os.popen(cmd, 'w', bufsize)
    # ==>
    pipe = Popen(cmd, shell=True, bufsize=bufsize, stdin=PIPE).stdin

Warning Use communicate() rather than
stdin.write(), or to avoid deadlocks due
to any of the other OS pipe buffers
filling up and blocking the child

So your example could be written as follows:

from subprocess import Popen, PIPE, STDOUT

p = Popen(['grep', 'f'], stdout=PIPE, stdin=PIPE, stderr=STDOUT)    
grep_stdout = p.communicate(input=b'one\ntwo\nthree\nfour\nfive\nsix\n')[0]
# -> four
# -> five
# ->

On Python 3.5+ (3.6+ for encoding), you could use, to pass input as a string to an external command and get its exit status, and its output as a string back in one call:

#!/usr/bin/env python3
from subprocess import run, PIPE

p = run(['grep', 'f'], stdout=PIPE,
        input='one\ntwo\nthree\nfour\nfive\nsix\n', encoding='ascii')
# -> 0
# -> four
# -> five
# -> 

This question is answered By – jfs

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0