我使用日期+“%T”打印开始和结束时间，结果如下:
10:33:56
10:36:10
我如何计算并打印这两者之间的差值呢?
我想要的是:
2m 14s
当前回答
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 $1}') seconds."
您将得到如下输出。
2012 10 16 01 56 37
2014 09 17 18 24 02
The sample time is 700 days, 16 hours, 27 minutes, and 25 seconds.
我修改了一点脚本，使其独立(即。只是设置变量值)，但也许总体思想也是如此。您可能需要对负值进行额外的错误检查。
其他回答
% start=$(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。
我想提出另一种避免召回日期命令的方法。如果你已经收集了%T日期格式的时间戳，这可能会有帮助:
ts_get_sec()
{
read -r h m s <<< $(echo $1 | tr ':' ' ' )
echo $(((h*60*60)+(m*60)+s))
}
start_ts=10:33:56
stop_ts=10:36:10
START=$(ts_get_sec $start_ts)
STOP=$(ts_get_sec $stop_ts)
DIFF=$((STOP-START))
echo "$((DIFF/60))m $((DIFF%60))s"
我们甚至可以用同样的方法处理毫秒。
ts_get_msec()
{
read -r h m s ms <<< $(echo $1 | tr '.:' ' ' )
echo $(((h*60*60*1000)+(m*60*1000)+(s*1000)+ms))
}
start_ts=10:33:56.104
stop_ts=10:36:10.102
START=$(ts_get_msec $start_ts)
STOP=$(ts_get_msec $stop_ts)
DIFF=$((STOP-START))
min=$((DIFF/(60*1000)))
sec=$(((DIFF%(60*1000))/1000))
ms=$(((DIFF%(60*1000))%1000))
echo "${min}:${sec}.$ms"
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。
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