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Display runnning countdown in a bash script?

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Top Forums Shell Programming and Scripting Display runnning countdown in a bash script?
# 8  
Old 04-02-2009
Quote:
Originally Posted by pvdb
I'm assuming doing it does way minimizes the "skew" on the actual time spent in the function, ie. every cycle is as close to 1 second as possible, as opposed to 1 second + however long it takes to do all the work.

Is that correct, or is there another benefit in doing it this way?

That's exactly right.
# 9  
Old 04-02-2009
Hi Corona688,

Quote:
Originally Posted by Corona688
Great idea, thanks for the patch!
All good points you raise... here's a unified diff of the two scripts, which also addresses the issue you correctly point out with printf control characters in the prefix:

Code:
pvdb@localhost ~$ diff -U 3 original prefix 
--- original	2009-04-02 22:08:47.000000000 +0100
+++ prefix	2009-04-02 22:13:06.000000000 +0100
@@ -2,7 +2,9 @@
 {
         local OLD_IFS="${IFS}"
         IFS=":"
-        local ARR=( $1 )
+        local ARR=( $1 ) ; shift
+        IFS="${OLD_IFS}"
+        local PREFIX="$*" ; [ -n "${PREFIX}" ] && PREFIX="${PREFIX} > "
         local SECONDS=$((  (ARR[0] * 60 * 60) + (ARR[1] * 60) + ARR[2]  ))
         local START=$(date +%s)
         local END=$((START + SECONDS))
@@ -13,12 +15,11 @@
                 CUR=$(date +%s)
                 LEFT=$((END-CUR))
 
-                printf "\r%02d:%02d:%02d" \
-                        $((LEFT/3600)) $(( (LEFT/60)%60)) $((LEFT%60))
+                printf "\r%s%02d:%02d:%02d" \
+                        "${PREFIX}" $((LEFT/3600)) $(( (LEFT/60)%60)) $((LEFT%60))
 
                 sleep 1
         done
-        IFS="${OLD_IFS}"
         echo "        "
 }
 
pvdb@localhost ~/Desktop $ _

... as well as the patched script in its entirity:

Code:
function countdown
{
        local OLD_IFS="${IFS}"
        IFS=":"
        local ARR=( $1 ) ; shift
        IFS="${OLD_IFS}"
        local PREFIX="$*" ; [ -n "${PREFIX}" ] && PREFIX="${PREFIX} > "
        local SECONDS=$((  (ARR[0] * 60 * 60) + (ARR[1] * 60) + ARR[2]  ))
        local START=$(date +%s)
        local END=$((START + SECONDS))
        local CUR=$START

        while [[ $CUR -lt $END ]]
        do
                CUR=$(date +%s)
                LEFT=$((END-CUR))

                printf "\r%s%02d:%02d:%02d" \
                        "${PREFIX}" $((LEFT/3600)) $(( (LEFT/60)%60)) $((LEFT%60))

                sleep 1
        done
        echo "        "
}

Regards,

PVDB
# 10  
Old 04-02-2009

That still fails if the arg includes 08 or 09.
# 11  
Old 04-02-2009
Hi cfajohnson,

Quote:
Originally Posted by cfajohnson

That still fails if the arg includes 08 or 09.
Yes, absolutely agree, and your version of the script is definitely the way to go, but by posting my corrected patch, I was just finishing what I started, so to speak Smilie
# 12  
Old 04-02-2009

I should have added that yours is easily fixed:

Code:
local SECONDS=$(( ${ARR[0]#0} * 3600 + ${ARR[1]#0} * 60 + ${ARR[2]#0} ))
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