如何杀死描述中包含'/var/'字符串的所有进程？

Question

I am using lsof to find out all processes which have /var/ in the description.

[root@localhost ~]# lsof | grep /var/ | less
systemd      1                          root  105u     unix 0xffff9b25fae49680      0t0      21311 /var/run/cups/cups.sock type=STREAM
systemd      1                          root  134u     unix 0xffff9b25f509cd80      0t0      21334 /var/run/libvirt/virtlogd-sock type=STREAM
systemd      1                          root  135u     unix 0xffff9b25fae49f80      0t0      21315 /var/run/.heim_org.h5l.kcm-socket type=STREAM
systemd      1                          root  137u     unix 0xffff9b25fae48d80      0t0      21318 /var/run/libvirt/virtlockd-sock type=STREAM
polkitd    745                       polkitd  mem       REG                8,3 10406312    2107847 /var/lib/sss/mc/initgroups
polkitd    745                       polkitd  mem       REG                8,3  6406312    2107846 /var/lib/sss/mc/group

I am now trying to kill all these processes using the following command:

kill -9 $(lsof | grep /var/)

But getting error:

-bash: kill: root: arguments must be process or job IDs
-bash: kill: 8w: arguments must be process or job IDs
-bash: kill: REG: arguments must be process or job IDs
-bash: kill: 8,3: arguments must be process or job IDs

Answer 1

You want to

• match on just the NAME field, and
• print out just the pid, not the whole line:

lsof | awk -v pattern="/var/" '$9 ~ pattern {print $2}'

On my system, that seems to output lots of duplicates:

lsof | awk -v pattern="/var/" '$9 ~ pattern {print $2}' | sort -u

and pass that list of pids to kill using xargs

lsof | awk -v pattern="/var/" '$9 ~ pattern {print $2}' | sort -u | xargs echo kill

Remove "echo" if you're satisfied it's working as you expect.