How do I create a stopwatch Bash script to constantly display elapsed time?

荒凉一梦 提交于 2019-12-05 14:09:29

I use the following code snippet in long-running scripts. The timer runs in a function (separate process) which will be killed (trap) if the main process is terminated by a keybord interrupt. The output shows the time elapsed from the start of the timer:

... working [hh:mm:ss]  00:07:58

The snippet:

#===  FUNCTION  ================================================================
#          NAME:  progressIndicatorTime
#   DESCRIPTION:  Display a progress indicator with seconds passed.
#    PARAMETERS:  [increment]   between 1 and 60 [sec], default is 2 [sec]
#===============================================================================
function progressIndicatorTime ()
{
  declare default=2                                       # default increment [sec]
  declare increment="${1:-$default}"                      # 1. parameter or default
  declare format='\b\b\b\b\b\b\b\b%02d:%02d:%02d'         # time format hh:mm:ss
  declare timepassed=0
  declare seconds minutes hours

  [[ ! "$increment" =~ ^([1-9]|[1-5][0-9]|60)$ ]] && increment=$default
  printf " ... working [hh:mm:ss]  00:00:00"
  while : ; do                                            # infinite loop 
    ((seconds=timepassed%60))
    ((minutes=timepassed/60))
    ((hours=minutes/60))
    ((minutes=minutes%60))
    printf "$format" $hours $minutes $seconds
    sleep $increment || break                             # sleep ...
    ((timepassed+=increment))
  done
}    # ----------  end of function progressIndicatorTime  ----------

progressIndicatorTime &                                   # run progress indicator
declare progressIndicatorPid=${!}                         # save process ID
trap  "kill $progressIndicatorPid" INT TERM               # trap keyboard interrupt

#
# run long command
#

kill -s SIGTERM $progressIndicatorPid                     # terminate progress indicator

The art could use some work, but give this a try:

#!/bin/bash

ref_date='Thu Apr 19 17:07:39 CDT 2012'
ref_sec=$(date -j -f '%a %b %d %T %Z %Y' "${ref_date}" +%s)
update_inc=1

tput clear
cat <<'EOF'


            [|]     [|]
         _-'''''''''''''-_
        /                 \
       |                   |
       |                   |
       |                   |
        \                 /
         '-_____________-'
EOF

while :
do
  ((sec=$(date +%s) - ${ref_sec}))
  ((day=sec/86400))
  ((sec-=day*86400))
  ((hour=sec/3600))
  ((sec-=hour*3600))
  ((min=sec/60))
  ((sec-=min*60))
  tput cup 6 14
  printf "%.2id:%.2ih:%.2im:%.2is\r" ${day} ${hour} ${min} ${sec}
  sleep ${update_inc}
done

exit 0

Note that the syntax of the first date command is for OSX.

For GNU date, use date --date="${ref_date}" +%s

If you have Bash4 or ksh93, you can use the printf %()T syntax to print strftime formats. The following example is Bash 4 printing in the format you want. Bash precision is limited to 1 second. ksh93 supports floats and would require some modification.

#!/usr/bin/env bash

# To supply a default date/time. To use now as the default, leave empty, or run with a null first arg.
deftime='Fri Apr 14 14:00:00 EDT 2011'

if (( ${BASH_VERSINFO[0]}${BASH_VERSINFO[1]} >= 42 )); then
    unixTime() {
        printf ${2+-v "$2"} '%(%s)T' -1
    }
else
    unixTime() {
        printf ${2+-v "$2"} "$(date '+%s')"
    }
fi

stopWatch() {
    local timestamp="$(date ${1:+'-d' "$1"} '+%s')" curtime day=$((60*60*24)) hour=$((60**2))
    # unixTime -1 timestamp

    while
        unixTime -1 curtime
        (( curtime -= timestamp ))
        printf '%02dd:%02dh:%02dm:%02ds\r' $(( curtime / day )) $(( (curtime / hour) % 24 )) $(( (curtime / 60) % 60 )) $(( curtime % 60 ))
        do sleep 1
    done
}

stopWatch "${1-deftime}"

You may also be interested in the $SECONDS variable. Unfortunately there is no convienient way to accept a human-readable date like you want. It would require considerable parsing effort. The date command (particularly GNU date) can do that to a limited extent.

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