centos7. 防火墙启动失败Job for firewalld.service failed because the control process exited with error code.

叁歲伎倆 2023-07-11 14:00 10阅读 0赞

systemctl status firewalld.service  
● firewalld.service - firewalld - dynamic firewall daemon  
   Loaded: loaded (/usr/lib/systemd/system/firewalld.service; enabled; vendor preset: enabled)  
   Active: failed (Result: exit-code) since Fri 2020-03-06 00:51:01 CST; 4min 40s ago  
     Docs: man:firewalld(1)  
  Process: 30360 ExecStart=/usr/sbin/firewalld --nofork --nopid $FIREWALLD\_ARGS (code=exited, status=1/FAILURE)  
 Main PID: 30360 (code=exited, status=1/FAILURE)

Job for firewalld.service failed because the control process exited with error code. See "systemctl status firewalld.service" and "journalctl -xe" for details.  
\[root@iz2zei0su36x4uwu1tkrr9z ~\]\# journalctl -xe  
\-- The start-up result is done.  
Mar 06 00:51:01 nameddd CROND\[30363\]: (root) CMD (/bin/bash /hom  
Mar 06 00:51:01 nameddd systemd\[1\]: Starting Session 2894 of use  
\-- Subject: Unit session-2894.scope has begun start-up  
\-- Defined-By: systemd  
\-- Support: http://lists.freedesktop.org/mailman/listinfo/systemd-devel  
\--  
\-- Unit session-2894.scope has begun starting up.

**原因：**

firewalld依赖于python，由于之前安装python3的时候，将默认python命令的别名指向了python3，造成无法启动

1、先找到firewalld的位置

\[root@dddd ~\]\# which firewalld  
    /usr/sbin/firewalld

find -name firewalld 这个不好使用which

2、将/usr/sbin/firewalld文件的第一行修改为旧的python2.7         如果还是不行试试python2、反正大意就是安装python错的。你得改回来。我改python2还是启动失败。centos7.4版本。

vim /usr/sbin/firewalld

\#!/usr/bin/python2.7 -Es  
      \# -\*- coding: utf-8 -\*-

将原来的python改为python2即可。

centos7. 防火墙启动失败Job for firewalld.service failed because the control process exited with error code.

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