how do I grep specific pattern from a Line using grep command?

to be more clear, I need to get the used space percentage (only the percentage without the other content in the line< of VOL (line 7) Using grep command*

2 Answers

Use grep with lookahead. It looks for one or more number followed by a % and only prints the number.

grep -Po "[0-9]+(?=%)" 

• -P : Perl-compatible regular expression so we can use lookahead.
• -o : Only print the matched parts.

$ echo '/dev/sda3 22G 4.1G 17G 20% /vol/store' | grep -Po "[0-9]+(?=%)" 20 

Another possibility:

echo "/dev/sda3 22G 4.1G 17G 20% /vol/store" | grep -o "[[:digit:]]*%" 20% 

Without the % sign:

echo "/dev/sda3 22G 4.1G 17G 20% /vol/store" | grep -Po "[[:digit:]]*(?=%)" 20