需要使用Shell脚本自动执行手动任务

Question

# find / -xdev -type f -perm -4000 -o -perm -2000 2>/dev/null -> To list out the full path of setuid program

After run the above command I will get the path of the program

/usr/bin/wall
/usr/bin/chfn
/usr/bin/chage
/usr/bin/gpasswd  
/usr/bin/newgrp
===================
So whatever the path it's coming, 

I need to add the rule "grep path /etc/audit/audit.rules" example : path /usr/bin/newgrp

  echo "-a always,exit -F path=/usr/bin/ssh-agent -F perm=x -F auid>=500 -F 
  auid!=4294967295 -k privileged" >>  /etc/audit/audit.rules  
  echo "-a always,exit -F path=/usr/bin/write -F perm=x -F auid>=500 -F 
  auid!=4294967295 -k privileged" >>  /etc/audit/audit.rules 
 echo "-a always,exit -F path=/usr/bin/locate -F perm=x -F auid>=500 -F 
 auid!=4294967295 -k privileged" >>  /etc/audit/audit.rules 
 echo "-a always,exit -F path=/usr/bin/wall -F perm=x -F auid>=500 -F 
auid!=4294967295 -k privileged" >>  /etc/audit/audit.rules

Please let me know is there anyway to automate the process without adding 
manually using echo. 

Thanks

Answer 1

Try this:

find / -xdev -type f -perm -4000 -o -perm -2000 | xargs -I {XXX} printf "'-a always,exit -F path='{XXX} ' -F perm=x -F auid>=500 -F auid!=4294967295 -k privileged'\n" >>/etc/audit/audit.rules

It pipes the find command which searches for set suid stuff to xargs which puts it's stuff from stdin (the stuff from find) in the corresponding place {XXX} in the printf.

Edit: took out usr/bin/