Match a time of day

A field takes a time like 09:30 or 23:59, and a plain digits pattern would also accept 99:99.

import re

rx = re.compile(r"([01][0-9]|2[0-3]):[0-5][0-9]")
for s in ["00:00", "09:30", "23:59", "24:00", "12:60", "9:30", "7:05"]:
    print(f"{s:6} {bool(rx.fullmatch(s))}")

relaxed = re.compile(r"([01]?[0-9]|2[0-3]):[0-5][0-9]")
print(bool(relaxed.fullmatch("9:30")), bool(relaxed.fullmatch("24:00")))

00:00  True
09:30  True
23:59  True
24:00  False
12:60  False
9:30   False
7:05   False
True False

The hour is either a 0 or 1 followed by any digit, giving 00 to 19, or a 2 followed by 0 to 3. The minute is a digit from 0 to 5 and any digit. The Python re documentation lists the syntax used. The first loop shows that 24:00 and 12:60 fail. The relaxed variant makes the first digit optional with ?, so 9:30 is accepted.

ISO 8601 allows 24:00 as the end of a day and a leap second as 23:59:60, and this pattern rejects both. Time zones, seconds and a 12 hour clock with AM or PM each need their own additions. If you only want valid times, parsing with datetime.strptime also checks the combination for you.