Regular expression to extract a percentage

Question

I have strings like the following: blabla a13724bla-bla244 35%

Notice that there is always a space before the percentage. I would like to extract the percentage number (so, without the %) from these strings using the Linux shell.

Answer 1

Assuming you have GNU grep:

$ grep -oP '\d+(?=%)' <<< "blabla a13724bla-bla244 35%"
35