2

I found here on superuser the following nice countdown script that I could include in my .bash_profile:

function countdown(){  
   date1=$((`date +%s` + $1));   
   while [ "$date1" -ne `date +%s` ]; do   
     echo -ne "$(date -u --date @$(($date1 - `date +%s`)) +%H:%M:%S)\r";  
     sleep 0.1  
  done  
}

It doesn't work on OSX, i understand, because of the differences in the date command. Another user commented as such but the problem remained unresolved in that question.

The same problem (I think) is also discussed and solved here but I cannot work out how to modify the countdown script to utilise this solution.

4

Personally, I'd refactor that a bit for readability (still relying on GNU date)

function countdown(){  
    local now=$(date +%s)
    local end=$((now + $1))
    while (( now < end )); do   
        printf "%s\r" "$(date -u -d @$((end - now)) +%T)"  
        sleep 0.25  
        now=$(date +%s)
    done  
    echo
}

Mapping the date calls to OSX's BSD date (man page here)

It looks like the GNU date invocation

date -u -d @$((end - now)) +%T

translates to this BSD date invocation

date -u -j -f %s $((end - now)) +%T

but that's untested.

| improve this answer | |
  • Bash is returning an unexpected EOF error on the printf line. – Tim Dec 9 '14 at 16:37
  • Are you sure you have all the quotes matching. The code I posted works (with bash version 4 at any rate) – glenn jackman Dec 9 '14 at 17:23
  • my fault: one of my double quotes was curly (damn autocorrect....) The code works perfectly now. – Tim Dec 10 '14 at 13:48
  • Time to pick a better editor. – glenn jackman Dec 10 '14 at 14:45
  • 1
    You could add if (( $# <1 )); then return; fi as the first line for a bit of robustness. – Burhan Ali Feb 2 '19 at 9:45

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