subprocess.check_output return code

Question

I am using:

grepOut = subprocess.check_output(\"grep \" + search + \" tmp\", shell=True)

To run a terminal command, I know that I can use a

Answer 1

is there a way to get a return code without a try/except?

check_output raises an exception if it receives non-zero exit status because it frequently means that a command failed. grep may return non-zero exit status even if there is no error -- you could use .communicate() in this case:

from subprocess import Popen, PIPE

pattern, filename = 'test', 'tmp'
p = Popen(['grep', pattern, filename], stdin=PIPE, stdout=PIPE, stderr=PIPE,
          bufsize=-1)
output, error = p.communicate()
if p.returncode == 0:
   print('%r is found in %s: %r' % (pattern, filename, output))
elif p.returncode == 1:
   print('%r is NOT found in %s: %r' % (pattern, filename, output))
else:
   assert p.returncode > 1
   print('error occurred: %r' % (error,))

You don't need to call an external command to filter lines, you could do it in pure Python:

with open('tmp') as file:
    for line in file:
        if 'test' in line:
            print line,

If you don't need the output; you could use subprocess.call():

import os
from subprocess import call
try:
    from subprocess import DEVNULL # Python 3
except ImportError: # Python 2
    DEVNULL = open(os.devnull, 'r+b', 0)

returncode = call(['grep', 'test', 'tmp'], 
                  stdin=DEVNULL, stdout=DEVNULL, stderr=DEVNULL)