regex - 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 (CST)

And I want to extract the 10:26 part.

Anybody knows of a way of doing this only with bash - without using sed, awk, etc?

Like, in PHP I would use - not the best way, but it works - something like:

preg_match( ""(d{2}:d{2}) PM (CST)"", "US/Central - 10:26 PM (CST)", $matches );

Thanks for any help, even if the answer uses sed or awk

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+Kd{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+Kd{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}}}'