How to store subprocess call output to a file python

Question

i have a script in bash that i am running in python using call from subprocess. i want the script output (strings) to be stored in a file. i tried working on a code that is

import logging
logging.basicConfig(level=logging.DEBUG,
                    format='',
                    datefmt='',
                    filename='E:\FYP\FYPPP\AMAPT\log.txt',
                    filemode='w')=
console = logging.StreamHandler()
console.setLevel(logging.INFO)
formatter = logging.Formatter('')
console.setFormatter(formatter)
logging.getLogger('').addHandler(console)
logging.info(call("sh amapt.sh", shell=True))
logger1 = logging.getLogger('myapp.area1')
logger2 = logging.getLogger('myapp.area2')
logger1.debug('test.')
logger1.info('test1.')
logger2.warning('test2.')
logger2.error('test3.') 

But displays the output as an integer of my script. check the image here I want the yellow/green output text to be stored in file but it is storing 255 instead of this text from my script.

Guide me please.

Answer 1

call returns the return code of your script. You need to get the output of the process:

logging.info(check_output("sh amapt.sh"))

notes:

• if process fails you'll get an exception
• you don't need shell=True

To get the output no matter what you could use Popen instead for instance like this:

logging.info(Popen("sh amapt.sh",stdout=PIPE,stderr=STDOUT).stdout.read())

The added ,stderr=STDOUT allows to get standard error in the log as well.

or from python 3.5 you can use subprocess.run() instead of Popen

logging.info(run("sh amapt.sh",stdout=PIPE,stderr=STDOUT).stdout.read())