Get the first percent from a list in bash

Question

I have been trying to find a solution all day. I eventually came access a question here that provided two commands. They are:

echo blabla 20% a13724bla-bla244 35% | sed -e 's/[^%0-9 ]*//g;s/  */\n/g' | sed -n '/%/p'
echo blabla 20% a13724bla-bla244 35% | sed 's/.*[ \t][ \t]*\([0-9][0-9]*\)%.*/\1/'

The first is supposed to give all the percentages found in the string and the second should only be used if you are expecting one percentage. My string will have more than one so I have been trying to use the first one. However, it returns all numbers in the strings and n representing spaces.

When I try to use the second command, I get the last percentage in my list which i don't need. I need the first percentage. Any help with this would be greatly appreciated.

Answer 1

perl would be easier to use as it supports non-greedy quantifier

$ echo 'blabla 20% a13724bla-bla244 35%' | perl -pe 's/.*?(\d+%).*/$1/'
20%

• .*? minimally match any character
• (\d+%) the first number followed by % combination
• .* rest of the line
• $1 replace the line with text matched within ()