bash script for timer

I want to write a bash script to run a timer for 30 seconds and, after 30 seconds, it should hold for the next 30 seconds and again that timer will start for next 30 seconds and continue the same.

for ((i=30; i>0; i--)); do sleep 1 & print "$i \r" wait
done

Using the code above, I can run the timer for 30seconds, but I need to hold for the next 30 seconds and again need to run the timer for the next 30 seconds.

How can I do this?

same thing I can do it in java using this code

import java.util.Timer;
import java.util.TimerTask;
public class TestClass {
    static int count = 0;
    public static void main(String[] args)  {
        final TestClass test = new TestClass();
                Timer timer = new Timer();
                timer.schedule(new TimerTask() {
                    @Override
                    public void run() {
                        test.doStuff(); }
                }, 0, 1000);
    }
    public void doStuff() {
        if (count == 30) {
            try {
                Thread.sleep(30000);
                count = 0;
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println(++count);
    }
}

2 Answers

Wrap your loop into an infinite loop (i.e. a while loop without conditions), and add a sleep 30 in this outer loop.

while :
do ### < your ### timer ### here > sleep 30
done

I'd suggest you also remove the wait instruction and the & from "sleep 1 &". Also, print is not the right program to write to the terminal, use echo instead.

while :
do for ((i=30; i>0; i--)) do sleep 1 echo -n "$i " done sleep 30 echo ""
done

Be aware that this timer is not accurate as evaluating the loop instructions takes a (small, but non zero) amount of time. A solution using date would be preferable: see for example the countdown function here: .

I think you are looking for something like this:

#!/bin/bash
## Infinite loop
while :; do ## Run a sleep command for 30 seconds in the background sleep 30 & ## $! is the PID of the last backgrounded process, so of the sleep. ## Wait for it to finish. c=30 while [[ -e /proc/$! ]]; do printf '%s\r' "$(( --c ))" ## For more precision, comment the line below. This causes the ## script to wait for a second so it doesn't spam your CPU. ## If you need accuracy, comment the line out. Although, if you ## really need accuracy, don't use the shell for this. sleep 1 done
done

Alternatively, use date:

#!/bin/bash
waitTime=30;
## Infinite loop
while :; do startTime=$(date +%s) currentTime=$(date +%s) c=$waitTime; while [[ $((currentTime - startTime)) -lt $waitTime ]]; do printf '%s\r' "$(( --c ))" ## For more precision, comment the line below. This causes the ## script to wait for a second so it doesn't spam your CPU. ## If you need accuracy, comment the line out. Although, if you ## really need accuracy, don't use the shell for this. sleep 1 currentTime=$(date +%s) done c=0 echo "";
done

For even greater precision, don't store the date in a variable:

#!/bin/bash
waitTime=4;
while :; do startTime=$(date +%s) c=$waitTime; while [[ $(($(date +%s) - startTime)) -lt $waitTime ]]; do printf '%s\r' "$(( --c ))" sleep 1 done c=0
done