CentOS防火墙开启失败怎么解决
突然之间发现centos 7 的防火墙无法启动，且firewall-cmd命令运行报错，防火墙启动报错如下：
[root@localhost firewalld]# systemctl start firewalld.service Job for firewalld.service failed because the control process exited with error code. See "systemctl status firewalld.service" and "journalctl -xe" for details. [root@localhost firewalld]# 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 三 2020-01-08 10:43:48 CST; 10s ago Docs: man:firewalld(1) Process: 29630 ExecStart=/usr/sbin/firewalld --nofork --nopid $FIREWALLD_ARGS (code=exited, status=127) Main PID: 29630 (code=exited, status=127)
firewall-cmd命令运行报错如下：
[root@localhost yangl]# firewall-cmd Traceback (most recent call last): File "/usr/bin/firewall-cmd", line 24, in <module> from gi.repository import GObject ImportError: No module named gi.repository
究其原因是由于使用了自定义安装的python，并将自己安装的python作为默认python导致（此外，将系统默认的python升级为python3也可能会出现类似问题）,因此只需将firewalld和firewall-cmd第一行调用的python改为系统的python即可解决。
我的python链接方式如下：
[root@localhost yangl]# cd /usr/bin/ [root@localhost bin]# ll python* lrwxrwxrwx. 1 root root 16 9月 28 2018 python -> /usr/bin/python2 lrwxrwxrwx. 1 root root 39 9月 28 2018 python2 -> /share/soft/python-2.7.15/bin/python2.7 -rwxr-xr-x. 1 root root 7216 7月 13 2018 python2.7 ＃系统的python lrwxrwxrwx. 1 root root 9 9月 28 2018 python2_old -> python2.7 lrwxrwxrwx. 1 root root 36 3月 7 2018 python3 -> /share/soft/python-3.6.4/bin/python3 lrwxrwxrwx. 1 root root 7 9月 28 2018 python_old -> python2
因此，只需vi打开/usr/sbin/firewalld 和/usr/bin/firewall-cmd，将其中的第一行由#!/usr/bin/python -Es 改为 #!/usr/bin/python2.7 -Es 即可！
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