Extract substring using regexp in plain bash
Using pure bash :
$ cat file.txtUS/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}}}'
echo "US/Central - 10:26 PM (CST)" | sed -n "s/^.*-\s*\(\S*\).*$/\1/p"-n suppress printings substitute^.* anything at the beginning- up until the dash\s* any space characters (any whitespace character)\( start capture group\S* any non-space characters\) end capture group.*$ anything at the end\1 substitute 1st capture group for everything on linep print it