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?
The most common way not to capture something is using look-around assertions: Use it like this
grep -oP '[0-9][0-9](?=%)'
It's worth noting that GNU grep support the -P option to enable Perl compatible regex syntax, however it is not included with OS X. On Linux, it will be available by default. A workaround would be to use ack instead.
But I'd still recommend to use GNU grep on OS X by default. It can be installed on OSX using Homebrew with the command brew grep install
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.
You can use sed as an alternative:
sed -rn 's/(^.*)([[:digit:]]{2})(%.*$)/\2/p' <<< "12G 39G 24% /dev"
Enable regular expressions with -r or -E and then split the line into 3 sections represented through parenthesis. Substitute the line for the second section only and print.
Use awk:
awk '{print $3+0}'
The value you seek is in the third field, and adding a zero coerces the string to a number, so % is removed.
