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@piraz
Last active April 6, 2022 02:58
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Script that monitors a service running on systemd. If service is not running the script will try to start the service.
#!/bin/python
#
# Copyright 2016 Flavio Garcia
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
#
# Usage: monitor_process.py <service_name>
#
# Example(crontab, every 5 minutes):
# */5 * * * * /root/bin/monitor_service.py prosody > /dev/null 2>&1
#
import sys
import subprocess
class ServiceMonitor(object):
def __init__(self, service):
self.service = service
def is_active(self):
"""Return True if service is running"""
cmd = '/bin/systemctl status %s.service' % self.service
proc = subprocess.Popen(cmd, shell=True,stdout=subprocess.PIPE,encoding='utf8')
stdout_list = proc.communicate()[0].split('\n')
for line in stdout_list:
if 'Active:' in line:
if '(running)' in line:
return True
return False
def start(self):
cmd = '/bin/systemctl start %s.service' % self.service
proc = subprocess.Popen(cmd, shell=True,stdout=subprocess.PIPE)
proc.communicate()
if __name__ == '__main__':
# TODO: Show usage
monitor = ServiceMonitor(sys.argv[1])
if not monitor.is_active():
monitor.start()
@hacktarus
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@piraz, i only changed the line 35 because the list at line 36 needed it
don't need to change line 45 because proc.communicate() accept the original code

@piraz
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piraz commented Aug 18, 2020

Done

@simbuawsassociate
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@hacktarus You are a lifesaver, Its helped me to fix my issue too. Thanks a lot!

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