Shell Programming and Scripting

Please how can i display a 60 second active countdown timer in an echo message.

e.g.

Code:

#!/bin/ksh
for (( i=60; i>0; i--)); do
  sleep 1 &
  printf "  $i \r"
  wait
done

This work in ksh as well. For posix you probably have to use a while loop.

Hi

this is not the best solution, but you can improve it.

Code:

input=60
tic=`date +%S`
elap_time=0

while [ "$elap_time" -le "$input" ]; do

toc=`date +%S`
elap_time=`expr $toc - $tic`
done

echo "time elapsed"

EDIT: Sorry I don't answ to your question, becouse you want display it.

Bye

Works (almost) fine in bash, too. But I don't understand the use of the sleep command as a background process. Furthermore, the "done" messages caused by returning sleeps disturb the display :-(

Nevertheless, much simplier would be just the good ole

Code:

sleep 60; echo "\a"

BR,

pen

Hi Pen, the sleep in the background ensures that you use as little time as possible processing the other commands in the loop. Otherwise total sleep is sometimes a bit more than 60 seconds.
You'll get done messages if you execute it from the command line, not if you execute it from a script. Alternatively you can enclose it in parentheses like so:

Code:

(for (( i=60; i>0; i--)); do
  sleep 1 &
  printf "  $i \r"
  wait
done)

Then you won't get the pesky job status.
Or you can put it in a function like CFA suggests..
(BTW I just changed the \b\b for \r because now I realise the backspaces weren't working properly).
Indeed, but it is my understanding that the OP requested a countdown timer.

Code:

i=60;while [ $i -gt 0 ];do if [ $i -gt 9 ];then printf "\b\b$i";else  printf "\b\b $i";fi;sleep 1;i=`expr $i - 1`;done