使用GNU日期(可靠的Ubuntu 14.04 LTS)概括@nisetama的解决方案:

start=`date`
# <processing code>
stop=`date`
duration=`date -ud@$(($(date -ud"$stop" +%s)-$(date -ud"$start" +%s))) +%T`

echo $start
echo $stop
echo $duration

收益率:

Wed Feb 7 12:31:16 CST 2018
Wed Feb 7 12:32:25 CST 2018
00:01:09

这是我的bash实现(bit从其他SO;-)

function countTimeDiff() {
    timeA=$1 # 09:59:35
    timeB=$2 # 17:32:55

    # feeding variables by using read and splitting with IFS
    IFS=: read ah am as <<< "$timeA"
    IFS=: read bh bm bs <<< "$timeB"

    # Convert hours to minutes.
    # The 10# is there to avoid errors with leading zeros
    # by telling bash that we use base 10
    secondsA=$((10#$ah*60*60 + 10#$am*60 + 10#$as))
    secondsB=$((10#$bh*60*60 + 10#$bm*60 + 10#$bs))
    DIFF_SEC=$((secondsB - secondsA))
    echo "The difference is $DIFF_SEC seconds.";

    SEC=$(($DIFF_SEC%60))
    MIN=$((($DIFF_SEC-$SEC)%3600/60))
    HRS=$((($DIFF_SEC-$MIN*60)/3600))
    TIME_DIFF="$HRS:$MIN:$SEC";
    echo $TIME_DIFF;
}

$ countTimeDiff 2:15:55 2:55:16
The difference is 2361 seconds.
0:39:21

未测试，可能有bug。

下面是一个只使用日期命令功能的解决方案，使用“ago”，而不使用第二个变量来存储完成时间:

#!/bin/bash

# save the current time
start_time=$( date +%s.%N )

# tested program
sleep 1

# the current time after the program has finished
# minus the time when we started, in seconds.nanoseconds
elapsed_time=$( date +%s.%N --date="$start_time seconds ago" )

echo elapsed_time: $elapsed_time

这给:

$ ./time_elapsed.sh 
elapsed_time: 1.002257120

GNU单位:

$ units
2411 units, 71 prefixes, 33 nonlinear units
You have: (10hr+36min+10s)-(10hr+33min+56s)
You want: s
    * 134
    / 0.0074626866
You have: (10hr+36min+10s)-(10hr+33min+56s)
You want: min
    * 2.2333333
    / 0.44776119

% start=$(date +%s)
% echo "Diff: $(date -d @$(($(date +%s)-$start)) +"%M minutes %S seconds")"
Diff: 00 minutes 11 seconds

Bash有一个方便的SECONDS内建变量，用于跟踪自shell启动以来已经过的秒数。此变量在赋值时保留其属性，赋值后返回的值为自赋值后的秒数加上赋值。

因此，您可以在启动计时事件之前将SECONDS设置为0，在事件发生后读取SECONDS，并在显示之前进行时间算术。

#!/usr/bin/env bash

SECONDS=0
# do some work
duration=$SECONDS
echo "$(($duration / 60)) minutes and $(($duration % 60)) seconds elapsed."

由于这个解决方案不依赖于date +%s(这是一个GNU扩展)，所以它可以移植到Bash支持的所有系统。