prevent subprocess.Popen from displaying output in python

Question

So I am trying to store the output of a command into a variable. I do not want it to display output while running the command though...

The code I have right now is as follows...

def getoutput(*args):
    myargs=args
    listargs=[l.split(' ',1) for l in myargs]
    import subprocess
    output=subprocess.Popen(listargs[0], shell=False ,stdout=subprocess.PIPE)   
    out, error = output.communicate()
    return(out,error)


def main():

    a,b=getoutput("httpd -S")

if __name__ == '__main__':
    main()

If I put this in a file and execute it on the command line. I get the following output even though I do not have a print statement in the code. How can I prevent this, while still storing the output?

#python ./apache.py 
httpd: Could not reliably determine the server's fully qualified domain name, using xxx.xxx.xxx.xx for ServerName
Syntax OK

Francis Avila · Accepted Answer · 2012-12-24 16:58:08Z

32

What you are seeing is standard-error output, not standard-output output. Stderr redirection is controlled by the stderr constructor argument. It defaults to None, which means no redirection occurs, which is why you see this output.

Usually it's a good idea to keep stderr output since it aids debugging and doesn't affect normal redirection (e.g. | and > shell redirection won't capture stderr by default). However you can redirect it somewhere else the same way you do stdout:

sp = subprocess.Popen(listargs[0], shell=False,
    stdout=subprocess.PIPE, stderr=subprocess.PIPE)
output, error = sp.communicate()

Or you can just drop stderr:

devnull = open(os.devnull, 'wb') #python >= 2.4
sp = subprocess.Popen(listargs[0], shell=False,
    stdout=subprocess.PIPE, stderr=devnull)

#python 3.x:
sp = subprocess.Popen(listargs[0], shell=False
    stdout=subprocess.PIPE, stderr=subprocess.DEVNULL)

edited Dec 24, 2012 at 16:58

answered Dec 24, 2012 at 16:31

Francis Avila

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1 Comment

jtaylor Over a year ago

you should use os.devnull for portability to windows

oathead · Accepted Answer · 2012-12-24 16:25:16Z

3

You're catching stdout, but you're not catching stderr(standard error) which I think is where that message is coming from.

output=subprocess.Popen(listargs[0], shell=False ,stdout=subprocess.PIPE, stderr=STDOUT)

That will put anything from stderr into the same place as stdout.

answered Dec 24, 2012 at 16:25

oathead

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