Print only regular expression match with sed command – Debian

Your bracket expression contains an A-z range that does not work with your current collation rules. If you add LC_ALL=C before sed command, it won’t error out, but it will still make it a bad regex since A-z ASCII char range also matches some non-letter symbols. It makes sense to replace A-z0-9 with [:alnum:].

So, you can either fix the regex and use 's/^ *Time zone: ([[:alnum:]_/]*).*$/1/p' or just capture any non-whitespaces there instead:

sed -n 's/^ *Time zone: ([^ ]*).*/1/p'

Details:

• -n – suppresses default line output
• ^ *Time zone: ([^ ]*).* – finds a line that starts with zero or more spaces, then has Time zone: string, then any zero or more chars other than space are captured into Group 1 (with ([^ ]*)) and the rest of the line (with .*),
• 1 – replaces the match with Group 1 value
• p – prints the result of the successful substitution

See the online demo:

#!/bin/bash
s='Local time: Wed 2022-11-16 13:02:00 CET
           Universal time: Wed 2022-11-16 12:02:00 UTC
                 RTC time: Wed 2022-11-16 12:02:01
                Time zone: Europe/Rome (CET, +0100)
System clock synchronized: yes
              NTP service: inactive
          RTC in local TZ: no'
sed -n 's/^ *Time zone: ([^ ]*).*$/1/p' <<< "$s"

Output:

Europe/Rome