How to extract value from shell and regex

Question

I have a string "12G 39G 24% /dev" . I have to extract the value '24'. I have used the below regex

grep '[0-9][0-9]%' -o

But I am getting output as 24%. I want only 24 as output and don't want '%' character. How to modify the regex script to extract only 24 as value?

question from:https://stackoverflow.com/questions/65921999/how-to-extract-value-from-shell-and-regex

Answer 1

One option would be to just grep again for the digits:

grep -o '[0-9][0-9]%' | grep -o '[0-9][0-9]'

However, if you want to accomplish this with a single regex, you can use the following:

grep -Po '[0-9]{2}(?=%)'

Note the -P option in this case; vanilla grep doesn't seem to support the (?=%) "look-around" part.