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。

日期可以给你的差异和格式为您(OS X选项显示)

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) +%T
# 00:02:14

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) \
    +'%-Hh %-Mm %-Ss'
# 0h 2m 14s

某些字符串处理可以删除这些空值

date -ujf%s $(($(date -jf%T "10:36:10" +%s) - $(date -jf%T "10:33:56" +%s))) \
    +'%-Hh %-Mm %-Ss' | sed "s/[[:<:]]0[hms] *//g"
# 2m 14s

如果你把较早的时间放在前面，这是行不通的。如果你需要处理,改变$(($(日期 ...) - $( 日期……)))(echo $(美元日期 ...) - $( 公元前日期…)| | tr - d -)

截至目前(GNU coreutils) 7.4，你现在可以使用-d来做算术:

$ date -d -30days
Sat Jun 28 13:36:35 UTC 2014

$ date -d tomorrow
Tue Jul 29 13:40:55 UTC 2014

你可以使用的单位是天、年、月、小时、分钟和秒:

$ date -d tomorrow+2days-10minutes
Thu Jul 31 13:33:02 UTC 2014

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

这里有一些魔法:

time1=14:30
time2=$( date +%H:%M ) # 16:00
diff=$(  echo "$time2 - $time1"  | sed 's%:%+(1/60)*%g' | bc -l )
echo $diff hours
# outputs 1.5 hours

Sed将a:替换为要转换为1/60的公式。然后用bc计算时间

#!/bin/bash

START_TIME=$(date +%s)

sleep 4

echo "Total time elapsed: $(date -ud "@$(($(date +%s) - $START_TIME))" +%T) (HH:MM:SS)"

$ ./total_time_elapsed.sh 
Total time elapsed: 00:00:04 (HH:MM:SS)