Run subprocess and print output to logging

Solution 1:

You could try to pass the pipe directly without buffering the whole subprocess output in memory:

from subprocess import Popen, PIPE, STDOUT

process = Popen(command_line_args, stdout=PIPE, stderr=STDOUT)
with process.stdout:
    log_subprocess_output(process.stdout)
exitcode = process.wait() # 0 means success

where log_subprocess_output() could look like:

def log_subprocess_output(pipe):
    for line in iter(pipe.readline, b''): # b'\n'-separated lines
        logging.info('got line from subprocess: %r', line)

Solution 2:

I am sure that there is the way to do it without creating temporary file to store process output

You simply have to check for the documentation of Popen, in particular about stdout and stderr:

stdin, stdout and stderr specify the executed program’s standard input, standard output and standard error file handles, respectively. Valid values are PIPE, an existing file descriptor (a positive integer), an existing file object, and None. PIPE indicates that a new pipe to the child should be created. With the default settings of None, no redirection will occur; the child’s file handles will be inherited from the parent. Additionally, stderr can be STDOUT, which indicates that the stderr data from the child process should be captured into the same file handle as for stdout.

So you can see that you can either use a file object, or the PIPE value. This allows you to use the communicate() method to retrieve the output:

from StringIO import StringIO
process = subprocess.Popen(arguments, stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
output, error = process.communicate()
log_subprocess_output(StringIO(output))

I'd rewrite your code as:

import shlex
import logging
import subprocess
from StringIO import StringIO

def run_shell_command(command_line):
    command_line_args = shlex.split(command_line)

    logging.info('Subprocess: "' + command_line + '"')

    try:
        command_line_process = subprocess.Popen(
            command_line_args,
            stdout=subprocess.PIPE,
            stderr=subprocess.STDOUT,
        )

        process_output, _ =  command_line_process.communicate()

        # process_output is now a string, not a file,
        # you may want to do:
        # process_output = StringIO(process_output)
        log_subprocess_output(process_output)
    except (OSError, CalledProcessError) as exception:
        logging.info('Exception occured: ' + str(exception))
        logging.info('Subprocess failed')
        return False
    else:
        # no exception was raised
        logging.info('Subprocess finished')

    return True

Solution 3:

I was trying to achieve the same on check_call and check_ouput. I found this solution to be working.

import logging
import threading
import os
import subprocess

logging.basicConfig(format='%(levelname)s:%(message)s', level=logging.INFO)

class LogPipe(threading.Thread):

    def __init__(self, level):
        """Setup the object with a logger and a loglevel
        and start the thread
        """
        threading.Thread.__init__(self)
        self.daemon = False
        self.level = level
        self.fdRead, self.fdWrite = os.pipe()
        self.pipeReader = os.fdopen(self.fdRead)
        self.start()

    def fileno(self):
        """Return the write file descriptor of the pipe"""
        return self.fdWrite

    def run(self):
        """Run the thread, logging everything."""
        for line in iter(self.pipeReader.readline, ''):
            logging.log(self.level, line.strip('\n'))

        self.pipeReader.close()

    def close(self):
        """Close the write end of the pipe."""
        os.close(self.fdWrite)

   def write(self):
       """If your code has something like sys.stdout.write"""
       logging.log(self.level, message)

   def flush(self):
       """If you code has something like this sys.stdout.flush"""
       pass

After implementing it, I performed the below steps:

try:
    # It works on multiple handlers as well
    logging.basicConfig(handlers=[logging.FileHandler(log_file), logging.StreamHandler()])
    sys.stdout = LogPipe(logging.INFO)
    sys.stderr = LogPipe(logging.ERROR)
...
    subprocess.check_call(subprocess_cmd, stdout=sys.stdout, stderr=sys.stderr)
    export_output = subprocess.check_output(subprocess_cmd, stderr=sys.stderr)
...
finally:
    sys.stdout.close()
    sys.stderr.close()
    # It is neccessary to close the file handlers properly.
    sys.stdout = sys.__stdout__
    sys.stderr = sys.__stderr__
    logging.shutdown()
    os.remove(log_file)

Solution 4:

This worked for me:

from subprocess import Popen, PIPE, STDOUT

command = f"shell command with arguments"
process = Popen(command, shell=True, stdout=PIPE, stderr=STDOUT)

with process.stdout:
    for line in iter(process.stdout.readline, b''):
        print(line.decode("utf-8").strip())

With exception handling:

from subprocess import Popen, PIPE, STDOUT, CalledProcessError

command = f"shell command with arguments"
process = Popen(command, shell=True, stdout=PIPE, stderr=STDOUT)

with process.stdout:
    try:
        for line in iter(process.stdout.readline, b''):
            print(line.decode("utf-8").strip())
            
    except CalledProcessError as e:
        print(f"{str(e)}")