[regex] Extract substring using regexp in plain bash

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

If your string is

foo="US/Central - 10:26 PM (CST)"

then

echo "${foo}" | cut -d ' ' -f3

will do the job.

    echo "US/Central - 10:26 PM (CST)" | sed -n "s/^.*-\s*\(\S*\).*$/\1/p"

-n      suppress printing
s       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 line
p       print it

Quick 'n dirty, regex-free, low-robustness chop-chop technique

string="US/Central - 10:26 PM (CST)"
etime="${string% [AP]M*}"
etime="${etime#* - }"