Trying to open a python script from another python script using Popen throws Windows Error

Question

I am trying to open a python script from another python script using Popen in Python 2.7.

The two scripts are:

child.py: Takes 5 ints one by one and waits some time and prints its square

    import time
    for i in range(5):
        value = int(raw_input('Enter an integer: '))
        time.sleep(2)
        print "Its square is ", value*value

parent.py: Opens child.py and writes 5 ints to its stdin and prints its stdout

    import subprocess, time

    # following is the line of interest
    child_program = subprocess.Popen("child.py", 
                                     stdin=subprocess.PIPE,
                                     stdout=subprocess.PIPE,
                                     stderr=subprocess.PIPE)

    for i in range(5):
        child_program.stdin.write(str(i)+"\n")
        child_program.stdin.flush()
        a = child_program.stdout.readline()
        print [a]  # I put it in list just to see exact format

The following are the other substitutions that I used for the first argument of Popen constructor in parent.py after seeing many similar question in stackoverflow

    "./child.py": Same WindowsError is produced
    <full path>:  Same WindowsError is produced
    ["python", "child.py"]: Did not raise error but opened python (useless)

The WindowsError produced is:

    WindowsError: [Error 193] %1 is not a valid Win32 application

score 0 · Answer 1 · answered Apr 29 '16 at 07:54

First you have to Specify the type of file you are trying to open as a subprocess

 child_program = subprocess.Popen(['executable','child.py'], 
                                 stdin=subprocess.PIPE,
                                 stdout=subprocess.PIPE,
                                 stderr=subprocess.PIPE)

This should resolve your issue.

Te-jé Rodgers · Answer 2 · 2016-04-29T08:35:00.847

If you have the Python interpreter setup to handle *.py files, then you can just set shell=True in your Popen constructor:

child_program = subprocess.Popen("child.py",
                                 shell=True,
                                 stdin=subprocess.PIPE,
                                 stdout=subprocess.PIPE,
                                 stderr=subprocess.PIPE)

As others have pointed out, this method comes with a security warning, so you'd have to take care that the arguments you pass to Popen aren't maliciously constructed (for instance, if you're getting some of your arguments from user input): https://docs.python.org/3/library/subprocess.html#security-considerations

Otherwise, you'll need to tell Popen to use a Python executable to load the file, like this:

import sys

child_program = subprocess.Popen([sys.executable, "child.py"],
                                 stdin=subprocess.PIPE,
                                 stdout=subprocess.PIPE,
                                 stderr=subprocess.PIPE)

In this context, sys.executable will resolve to the full path of the Python binary that was used to launch your parent script.

If you need shell=True, look for something better. Please recognize this is a security risk! — MKesper, Apr 29 '16 at 08:20

score 0 · Answer 3 · edited May 23 '17 at 11:59

Popen with the default shell=False delegates to CreateProcess in Windows as its docs say. That API function can only run executable files (regardless of their extension).

You need to either

add the path to the interpreter as the leading argument
use machinery that would delegate to ShellExecute that would run a command line like above accoring to the file type association. E.g. specify shell=True (heed the security warnings in the doc) because cmd.exe falls back to ShellExecute for non-executable files.

Trying to open a python script from another python script using Popen throws Windows Error

3 Answers3