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

#!/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)

下面是Daniel Kamil Kozar的回答，表示小时/分钟/秒:

echo "Duration: $(($DIFF / 3600 )) hours $((($DIFF % 3600) / 60)) minutes $(($DIFF % 60)) seconds"

所以完整的脚本是:

date1=$(date +"%s")
date2=$(date +"%s")
DIFF=$(($date2-$date1))
echo "Duration: $(($DIFF / 3600 )) hours $((($DIFF % 3600) / 60)) minutes $(($DIFF % 60)) seconds"

我知道这是一篇较老的文章，但我今天在编写一个脚本时偶然发现了它，该脚本将从日志文件中获取日期和时间并计算delta。下面的脚本当然是多余的，我强烈建议检查我的逻辑和数学。

#!/bin/bash

dTime=""
tmp=""

#firstEntry="$(head -n 1 "$LOG" | sed 's/.*] \([0-9: -]\+\).*/\1/')"
firstEntry="2013-01-16 01:56:37"
#lastEntry="$(tac "$LOG" | head -n 1 | sed 's/.*] \([0-9: -]\+\).*/\1/')"
lastEntry="2014-09-17 18:24:02"

# I like to make the variables easier to parse
firstEntry="${firstEntry//-/ }"
lastEntry="${lastEntry//-/ }"
firstEntry="${firstEntry//:/ }"
lastEntry="${lastEntry//:/ }"

# remove the following lines in production
echo "$lastEntry"
echo "$firstEntry"

# compute days in last entry
for i in `seq 1 $(echo $lastEntry|awk '{print $2}')`; do {
  case "$i" in
   1|3|5|7|8|10|12 )
    dTime=$(($dTime+31))
    ;;
   4|6|9|11 )
    dTime=$(($dTime+30))
    ;;
   2 )
    dTime=$(($dTime+28))
    ;;
  esac
} done

# do leap year calculations for all years between first and last entry
for i in `seq $(echo $firstEntry|awk '{print $1}') $(echo $lastEntry|awk '{print $1}')`; do {
  if [ $(($i%4)) -eq 0 ] && [ $(($i%100)) -eq 0 ] && [ $(($i%400)) -eq 0 ]; then {
    if [ "$i" = "$(echo $firstEntry|awk '{print $1}')" ] && [ $(echo $firstEntry|awk '{print $2}') -lt 2 ]; then {
      dTime=$(($dTime+1))
    } elif [ $(echo $firstEntry|awk '{print $2}') -eq 2 ] && [ $(echo $firstEntry|awk '{print $3}') -lt 29 ]; then {
      dTime=$(($dTime+1))
    } fi
  } elif [ $(($i%4)) -eq 0 ] && [ $(($i%100)) -ne 0 ]; then {
    if [ "$i" = "$(echo $lastEntry|awk '{print $1}')" ] && [ $(echo $lastEntry|awk '{print $2}') -gt 2 ]; then {
      dTime=$(($dTime+1))
    } elif [ $(echo $lastEntry|awk '{print $2}') -eq 2 ] && [ $(echo $lastEntry|awk '{print $3}') -ne 29 ]; then {
      dTime=$(($dTime+1))
    } fi
  } fi
} done

# substract days in first entry
for i in `seq 1 $(echo $firstEntry|awk '{print $2}')`; do {
  case "$i" in
   1|3|5|7|8|10|12 )
    dTime=$(($dTime-31))
    ;;
   4|6|9|11 )
    dTime=$(($dTime-30))
    ;;
   2 )
    dTime=$(($dTime-28))
    ;;
  esac
} done

dTime=$(($dTime+$(echo $lastEntry|awk '{print $3}')-$(echo $firstEntry|awk '{print $3}')))

# The above gives number of days for sample. Now we need hours, minutes, and seconds
# As a bit of hackery I just put the stuff in the best order for use in a for loop
dTime="$(($(echo $lastEntry|awk '{print $6}')-$(echo $firstEntry|awk '{print $6}'))) $(($(echo $lastEntry|awk '{print $5}')-$(echo $firstEntry|awk '{print $5}'))) $(($(echo $lastEntry|awk '{print $4}')-$(echo $firstEntry|awk '{print $4}'))) $dTime"
tmp=1
for i in $dTime; do {
  if [ $i -lt 0 ]; then {
    case "$tmp" in
     1 )
      tmp="$(($(echo $dTime|awk '{print $1}')+60)) $(($(echo $dTime|awk '{print $2}')-1))"
      dTime="$tmp $(echo $dTime|awk '{print $3" "$4}')"
      tmp=1
      ;;
     2 )
      tmp="$(($(echo $dTime|awk '{print $2}')+60)) $(($(echo $dTime|awk '{print $3}')-1))"
      dTime="$(echo $dTime|awk '{print $1}') $tmp $(echo $dTime|awk '{print $4}')"
      tmp=2
      ;;
     3 )
      tmp="$(($(echo $dTime|awk '{print $3}')+24)) $(($(echo $dTime|awk '{print $4}')-1))"
      dTime="$(echo $dTime|awk '{print $1" "$2}') $tmp"
      tmp=3
      ;;
    esac
  } fi
  tmp=$(($tmp+1))
} done

echo "The sample time is $(echo $dTime|awk '{print $4}') days, $(echo $dTime|awk '{print $3}') hours, $(echo $dTime|awk '{print $2}') minutes, and $(echo $dTime|awk '{print $1}') seconds."

您将得到如下输出。

2012 10 16 01 56 37
2014 09 17 18 24 02
The sample time is 700 days, 16 hours, 27 minutes, and 25 seconds.

我修改了一点脚本，使其独立(即。只是设置变量值)，但也许总体思想也是如此。您可能需要对负值进行额外的错误检查。

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。

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