Write a program called thetime that displays the time of day in am or pm notation rather than in 24-hour clock time.

Question

$ date
Wed Aug 28 19:34:01 EDT 2002
$ thetime
7:21 pm
Suggestion: Use the shell's built-in integer arithmetic to convert from 24-hour clock time. Then rewrite the program to use a case command instead. Rewrite it again to perform arithmetic with the expr command.

Answer 1

There are two ways to solve this. The date command has a rich formatting language that students can learn about by typing man strftime, which yields this delightful one-line command:
date +"%l:%M %p"
Failing that, the suggested solution produces a complex script:
hour=$(date | cut -d\ -f5 | cut -d: -f1)
min=$(date | cut -d\ -f5 | cut -d: -f2)

case $hour in
0[1-9]|[1-9]|10|11|12 ) echo "${hour}:$min am"
;;
0 ) echo "12:$min pm"
;;
* ) echo "$(( $hour - 12)):$min pm"
;;
esac
It's worth noting is that the date command reports hours in a 0-23 format, not 1-24, so midnight is not hour 24 on its 24 hour clock, but hour zero. That's why you need the special case for that situation so you don't get "-12:05 pm" as a late-night output.