Bash script to calculate time elapsed

Question

I am writing a script in bash to calculate the time elapsed for the execution of my commands, consider:

STARTTIME=$(date +%s)
#command block that takes time to c         


        

Answer 1

start=$(date +%Y%m%d%H%M%S);
for x in {1..5};
do echo $x;
sleep 1; done;
end=$(date +%Y%m%d%H%M%S);
elapsed=$(($end-$start));
ftime=$(for((i=1;i<=$((${#end}-${#elapsed}));i++));
        do echo -n "-";
        done;
        echo ${elapsed});
echo -e "Start  : ${start}\nStop   : ${end}\nElapsed: ${ftime}"

Start  : 20171108005304
Stop   : 20171108005310
Elapsed: -------------6

Answer 2

For larger numbers we may want to print in a more readable format. The example below does same as other but also prints in "human" format:

secs_to_human() {
    if [[ -z ${1} || ${1} -lt 60 ]] ;then
        min=0 ; secs="${1}"
    else
        time_mins=$(echo "scale=2; ${1}/60" | bc)
        min=$(echo ${time_mins} | cut -d'.' -f1)
        secs="0.$(echo ${time_mins} | cut -d'.' -f2)"
        secs=$(echo ${secs}*60|bc|awk '{print int($1+0.5)}')
    fi
    echo "Time Elapsed : ${min} minutes and ${secs} seconds."
}

Simple testing:

secs_to_human "300"
secs_to_human "305"
secs_to_human "59"
secs_to_human "60"
secs_to_human "660"
secs_to_human "3000"

Output:

Time Elapsed : 5 minutes and 0 seconds.
Time Elapsed : 5 minutes and 5 seconds.
Time Elapsed : 0 minutes and 59 seconds.
Time Elapsed : 1 minutes and 0 seconds.
Time Elapsed : 11 minutes and 0 seconds.
Time Elapsed : 50 minutes and 0 seconds.

To use in a script as described in other posts (capture start point then call the function with the finish time:

start=$(date +%s)
# << performs some task here >>
secs_to_human "$(($(date +%s) - ${start}))"

Answer 3

I find it very clean to use the internal variable "$SECONDS"

SECONDS=0 ; sleep 10 ; echo $SECONDS

Answer 4

Try the following code:

start=$(date +'%s') && sleep 5 && echo "It took $(($(date +'%s') - $start)) seconds"

Answer 5

This is a one-liner alternative to Mike Q's function:

secs_to_human() {
    echo "$(( ${1} / 3600 ))h $(( (${1} / 60) % 60 ))m $(( ${1} % 60 ))s"
}

Answer 6

You can use Bash's time keyword here with an appropriate format string

TIMEFORMAT='It takes %R seconds to complete this task...'
time {
    #command block that takes time to complete...
    #........
 }

Here's what the reference says about TIMEFORMAT:

The value of this parameter is used as a format string specifying how the timing information for pipelines prefixed with the time reserved word should be displayed. The ‘%’ character introduces an escape sequence that is expanded to a time value or other information. The escape sequences and their meanings are as follows; the braces denote optional portions.

%%

    A literal ‘%’.
%[p][l]R

    The elapsed time in seconds.
%[p][l]U

    The number of CPU seconds spent in user mode.
%[p][l]S

    The number of CPU seconds spent in system mode.
%P

    The CPU percentage, computed as (%U + %S) / %R. 

The optional p is a digit specifying the precision, the number of fractional digits after a decimal point. A value of 0 causes no decimal point or fraction to be output. At most three places after the decimal point may be specified; values of p greater than 3 are changed to 3. If p is not specified, the value 3 is used.

The optional l specifies a longer format, including minutes, of the form MMmSS.FFs. The value of p determines whether or not the fraction is included.

If this variable is not set, Bash acts as if it had the value

$'\nreal\t%3lR\nuser\t%3lU\nsys\t%3lS'

If the value is null, no timing information is displayed. A trailing newline is added when the format string is displayed.