Output from subprocess.Popen

Question

I\'ve been writing some python code and in my code I was using \"command\"

The code was working as I intended but then I noticed in the Python docs that command has

Answer 1

The "ugly string" is what it should be printing. Python is correctly printing out the repr(subprocess.Popen(...)), just like what it would print if you said print(open('myfile.txt')).

Furthermore, python has no knowledge of what is being output to stdout. The output you are seeing is not from python, but from the process's stdout and stderr being redirected to your terminal as spam, that is not even going through the python process. It's like you ran a program someprogram & without redirecting its stdout and stderr to /dev/null, and then tried to run another command, but you'd occasionally see spam from the program. To repeat and clarify:

  <-- output of python program
brettg@underworld:~/dev$ total 52                     <-- spam from your shell, not from python
drwxr-xr-x  3 brettg brettg 4096 2011-05-27 12:38 .   <-- spam from your shell, not from python
drwxr-xr-x 21 brettg brettg 4096 2011-05-24 17:40 ..  <-- spam from your shell, not from python
...

In order to capture stdout, you must use the .communicate() function, like so:

#!/usr/bin/python
import subprocess
output = subprocess.Popen(["ls", "-a", "-l"], stdout=subprocess.PIPE).communicate()[0]
print output

Furthermore, you never want to use shell=True, as it is a security hole (a major security hole with unsanitized inputs, a minor one with no input because it allows local attacks by modifying the shell environment). For security reasons and also to avoid bugs, you generally want to pass in a list rather than a string. If you're lazy you can do "ls -al".split(), which is frowned upon, but it would be a security hole to do something like ("ls -l %s"%unsanitizedInput).split().