Shell Function not Executing within IF clause

Question

I was writing a code to print dots for every 10 seconds of sleep. Also, at every sleep session end, the sleep interval should increase by 5. And if remaining sleep sessions is less than half of total sleep sessions (i.e. 6) then the script should display the elapsed time.

Here is the code,

#!/bin/bash

current_sleep_sess=0
total_sleep_sess=6
rem_sleep_sess=6
sleep_interval=10
SECONDS=0

elapsed_time()
{
    sec=$1
    (( sec < 60 )) && echo -e "[Elapsed Time : $sec sec]\c"

    (( sec >= 60 && sec > 3600 )) && echo -e "[$(( sec / 60 )) min $(( sec % 60 )) sec]\c"

    (( sec >= 3600 )) && echo -e "[$(( sec / 3600 )) hrs $(( (sec % 3600) / 60)) min $(( (sec % 3600) % 60 )) sec]\c"
}

until (( rem_sleep_sess == 0 ))
do

    (( current_sleep_sess=$current_sleep_sess+1 ))

    sleep $sleep_interval
    echo -e ".\c"

    if (( rem_sleep_sess <= $((total_sleep_sess / 2)) ))
    then
        elapsed_time $SECONDS
        echo
    fi

    (( sleep_interval=$sleep_interval+5 ))
    (( rem_sleep_sess=$total_sleep_sess-$current_sleep_sess ))

done

The problem is that when I use elapsed_time $SECONDS without the if clause, then the function executes, but when I use it within the if clause, then the function gets skipped. Is there something, some syntax rule that I am missing?

Michael Jaros · Accepted Answer · 2015-04-05 21:57:17Z

1

You wrote:

(( sec >= 60 && sec > 3600 )) && # [...]

but you probably meant:

(( sec >= 60 && sec < 3600 )) && # [...]

Output:

....[1 min 10 sec]
.[1 min 40 sec]
.[2 min 15 sec]

edited Apr 5, 2015 at 21:57

answered Apr 5, 2015 at 21:41

Michael Jaros

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