When I start apache then it gives me below error -
Job for apache2.service failed because the control process exited with error code. See "systemctl status apache2.service" and "journalctl -xe" for details.
I executed below command -
systemctl status apache2.service
I got below error
apache2.service - LSB: Apache2 web server
Loaded: loaded (/etc/init.d/apache2; bad; vendor preset: enabled)
Drop-In: /lib/systemd/system/apache2.service.d
└─apache2-systemd.conf
Active: failed (Result: exit-code) since Wed 2017-05-31 17:01:40 IST; 1min 34s ago
Docs: man:systemd-sysv-generator(8)
Process: 11884 ExecStop=/etc/init.d/apache2 stop (code=exited, status=0/SUCCESS)
Process: 3171 ExecReload=/etc/init.d/apache2 reload (code=exited, status=0/SUCCESS)
Process: 12696 ExecStart=/etc/init.d/apache2 start (code=exited, status=1/FAILURE)
May 31 17:01:40 itadmin systemd[1]: Starting LSB: Apache2 web server...
May 31 17:01:40 itadmin apache2[12696]: * Starting Apache httpd web server apache2
May 31 17:01:40 itadmin apache2[12696]: mktemp: failed to create file via template ‘/tmp/tmp.XXXXXXXXXX’: Read-only file system
May 31 17:01:40 itadmin apache2[12696]: /etc/init.d/apache2: 72: /etc/init.d/apache2: cannot create : Directory nonexistent
May 31 17:01:40 itadmin apache2[12696]: *
May 31 17:01:40 itadmin apache2[12696]: * The apache2 configtest failed.
May 31 17:01:40 itadmin systemd[1]: apache2.service: Control process exited, code=exited status=1
May 31 17:01:40 itadmin systemd[1]: Failed to start LSB: Apache2 web server.
May 31 17:01:40 itadmin systemd[1]: apache2.service: Unit entered failed state.
May 31 17:01:40 itadmin systemd[1]: apache2.service: Failed with result 'exit-code'.
I am not understanding anything. Please anyone can help me?