以下是我的做法:

START=$(date +%s);
sleep 1; # Your stuff
END=$(date +%s);
echo $((END-START)) | awk '{print int($1/60)":"int($1%60)}'

非常简单，取开始时的秒数，然后取结束时的秒数，打印出以分钟为单位的差值:秒。

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。

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。

我需要一个时差脚本用于mencoder(它的——endpos是相对的)，我的解决方案是调用一个Python脚本:

$ ./timediff.py 1:10:15 2:12:44
1:02:29

还支持秒的分数:

$ echo "diff is `./timediff.py 10:51.6 12:44` (in hh:mm:ss format)"
diff is 0:01:52.4 (in hh:mm:ss format)

它可以告诉你200和120的差值是1h 20m:

$ ./timediff.py 120:0 200:0
1:20:0

并且可以将任何(可能是分数)秒、分或小时数转换为hh:mm:ss

$ ./timediff.py 0 3600
1:00:0
$ ./timediff.py 0 3.25:0:0
3:15:0

timediff.py:

#!/usr/bin/python

import sys

def x60(h,m):
    return 60*float(h)+float(m)

def seconds(time):
    try:
       h,m,s = time.split(':')
       return x60(x60(h,m),s)
    except ValueError:
       try:
          m,s = time.split(':')
          return x60(m,s)
       except ValueError:
          return float(time)

def difftime(start, end):
    d = seconds(end) - seconds(start)
    print '%d:%02d:%s' % (d/3600,d/60%60,('%02f' % (d%60)).rstrip('0').rstrip('.'))

if __name__ == "__main__":
   difftime(sys.argv[1],sys.argv[2])

以下是我的做法:

START=$(date +%s);
sleep 1; # Your stuff
END=$(date +%s);
echo $((END-START)) | awk '{print int($1/60)":"int($1%60)}'

非常简单，取开始时的秒数，然后取结束时的秒数，打印出以分钟为单位的差值:秒。

#!/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)