Extract substring using regexp in plain bash

Question

I\'m trying to extract the time from a string using bash, and I\'m having a hard time figuring it out.

My string is like this:

US/Central - 10:26 PM          


        

Answer 1

Using pure bash :

$ cat file.txt
US/Central - 10:26 PM (CST)
$ while read a b time x; do [[ $b == - ]] && echo $time; done < file.txt

another solution with bash regex :

$ [[ "US/Central - 10:26 PM (CST)" =~ -[[:space:]]*([0-9]{2}:[0-9]{2}) ]] &&
    echo ${BASH_REMATCH[1]}

another solution using grep and look-around advanced regex :

$ echo "US/Central - 10:26 PM (CST)" | grep -oP "\-\s+\K\d{2}:\d{2}"

another solution using sed :

$ echo "US/Central - 10:26 PM (CST)" |
    sed 's/.*\- *\([0-9]\{2\}:[0-9]\{2\}\).*/\1/'

another solution using perl :

$ echo "US/Central - 10:26 PM (CST)" |
    perl -lne 'print $& if /\-\s+\K\d{2}:\d{2}/'

and last one using awk :

$ echo "US/Central - 10:26 PM (CST)" |
    awk '{for (i=0; i<=NF; i++){if ($i == "-"){print $(i+1);exit}}}'