Grep output help needed


 
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# 1  
Old 05-23-2011
Grep output help needed

According to linux hardening guide a PATH variable must not contain . or ..
so i grep the path variable using.

Code:
# echo $PATH | grep "."
/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/root/bin
# echo $PATH | grep ".."
/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/root/bin

There are no . or .. in the PATH but still the output is produced.

I m little confused How its possible?
Code:
# echo $PATH
/usr/kerberos/sbin:/usr/kerberos/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/root/bin
# pwd
/usr/local/sbin

What logic can be applied to check the existence of . and .. in PATH variable?
# 2  
Old 05-23-2011
In regular expression '.' stands for any character. So you are trying to grep one char or two chars.

Try:

Code:
echo $PATH|grep "\."

echo $PATH|grep "\.\."

These 2 Users Gave Thanks to kumaran_5555 For This Post:
# 3  
Old 05-23-2011
u can also try this

echo $PATH | grep '.'
These 2 Users Gave Thanks to atul9806 For This Post:
 
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