Python 脚本，用于监视服务的 systemctl 状态并在返回 false 时通过 sendmail 发送电子邮件

Question

这里有点菜鸟。

我正在尝试编写一个 python 脚本，该脚本本质上将使用 systemctl 实用程序查看服务的子状态，然后如果检测到子状态不是“正在运行”，则使用 sendmail 发送电子邮件。

对于这个特定的实例，仅仅显示服务是否处于事件状态是不够的，还需要显示服务是否正在运行。

我有一个一直在使用的脚本，但它最初是在查找 URL 状态，我将其更改为查看各个服务。我希望监控 6 个不同的服务，但如果其中任何一个服务返回除“正在运行”之外的任何内容，请发送电子邮件

下面是我到目前为止的脚本:

#!/usr/bin/python


import time
import os
from email.mime.text import MIMEText
from subprocess import Popen, PIPE

#The url being monitored.
service1 = "tomcat"
#service2 = "hostcontext"
#service3 = "ecs-ec"
#service4 = "ecs-ep"
#service5 = "ariel_proxy_server"
#service6 = "ecs-ec-ingress"


#The contents of the email
msg = MIMEText(service1 + " is not responding.  Please investigate.")
msg["From"] = "sending address"
msg["To"] = "recipient address"
msg["Subject"] = service1 + "is not responding"

#This loops while the script is running.
# It gets the status returned from the urllib call, if it's not 200 it will email the email contents above.
while True:
    status = os.system('systemctl show tomcat -p SubState | sed "s/SubState=//g"')

    if status == 'running':
        #This is what sends the email.  If you don't have sendmail then update this.
        p = Popen(["/usr/sbin/sendmail", "-t", "-oi"], stdin=PIPE)
        p.communicate(msg.as_string())
    #The number of seconds the loop will pause for before checking again.  I set it to 60.
    time.sleep(60)

Answer 1

我会做这样的事情:

import time, traceback
from subprocess import getoutput
services = ["mariadb", "apache"]

while 1:
    for service in services:
        try:
            status = getoutput(f"systemctl show {service} -p SubState").split("=")[1].lower()
            if status != 'running':
                print(f"Service {service} is down, send email...")
        except:
            pass
            print(traceback.print_exc())
            # log the error if needed
    time.sleep(60)