Bash脚本：两次之间的分钟差异

I have two time strings; eg. "09:11" and "17:22" on the same day (format is hh:mm). How do I calculate the time difference in minutes between these two?

我有两个时间字符串;例如。同一天“09:11”和“17:22”（格式为hh：mm）。如何计算这两者之间的分钟时差？

Can the standard date library do this?

标准日期库可以这样做吗？

Example:

例：

#!/bin/bash

MPHR=60    # Minutes per hour.

CURRENT=$(date -u -d '2007-09-01 17:30:24' '+%F %T.%N %Z')
TARGET=$(date -u -d'2007-12-25 12:30:00' '+%F %T.%N %Z')

MINUTES=$(( $(diff) / $MPHR ))

Is there a simpler way of doing this given the hour and minute in hh:mm

考虑到hh：mm中的小时和分钟，有没有更简单的方法

7 个解决方案

#1

A pure bash solution :

纯粹的bash解决方案：

old=09:11
new=17:22

# feeding variables by using read and splitting with IFS
IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"

# convert hours to minutes
# the 10# is there to avoid errors with leading zeros
# by telling bash that we use base 10
total_old_minutes=$((10#$old_hour*60 + 10#$old_min))
total_minutes=$((10#$hour*60 + 10#$min))

echo "the difference is $((total_minutes - total_old_minutes)) minutes"

Another solution using date (we work with hour/minutes, so the date is not important)

使用日期的另一个解决方案（我们使用小时/分钟，因此日期并不重要）

old=09:11
new=17:22

IFS=: read old_hour old_min <<< "$old"
IFS=: read hour min <<< "$new"

# convert the date "1970-01-01 hour:min:00" in seconds from Unix EPOCH time
sec_old=$(date -d "1970-01-01 $old_hour:$old_min:00" +%s)
sec_new=$(date -d "1970-01-01 $hour:$min:00" +%s)

echo "the difference is $(( (sec_new - sec_old) / 60)) minutes"

See http://en.wikipedia.org/wiki/Unix_time

见http://en.wikipedia.org/wiki/Unix_time

#2

I would convert the dates to UNIX timestamps; you can subtract to get the difference in seconds, then divide by 60:

我会将日期转换为UNIX时间戳;你可以减去以秒为单位得到差异，然后除以60：

#!/bin/bash

MPHR=60    # Minutes per hour.

CURRENT=$(date +%s -d '2007-09-01 17:30:24')
TARGET=$(date +%s -d'2007-12-25 12:30:00')

MINUTES=$(( ($TARGET - $CURRENT) / $MPHR ))

#3

MPHR=60
CURRENT=09:11
TARGET=17:22
echo $(( ( 10#${TARGET:0:2} - 10#${CURRENT:0:2} ) * MPHR + 10#${TARGET:4} - 10#${CURRENT:4} ))

#4

Here is how I did it:

我是这样做的：

START=$(date +%s);
sleep 1; # Your stuff
END=$(date +%s);
echo $((END-START)) | awk '{printf "%d:%02d:%02d", $1/3600, ($1/60)%60, $1%60}'

Really simple, take the number of seconds at the start, then take the number of seconds at the end, and print the difference in minutes:seconds.

非常简单，在开始时采用秒数，然后在结束时采用秒数，并以分钟为单位打印差异：秒。

#5

STARTTIME=$(date +%s)

YOUR CODES :

你的代码：

ENDTIME=$(date +%s)
secs=$(($ENDTIME - $STARTTIME))
printf 'Elapsed Time %dh:%dm:%ds\n' $(($secs/3600)) $(($secs%3600/60)) $(($secs%60))

#6

I was looking for a solution with seconds. Found here: How to calculate time difference in bash script?

我一直在寻找一个解决方案。在这里找到：如何计算bash脚本的时差？

#!/bin/bash
string1="10:33:56"
string2="10:36:10"
StartDate=$(date -u -d "$string1" +"%s")
FinalDate=$(date -u -d "$string2" +"%s")
date -u -d "0 $FinalDate sec - $StartDate sec" +"%H:%M:%S"

Here I have added seconds to Gilles' solution:

在这里，我为Gilles的解决方案增加了几秒钟：

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;
}

#7

@Dorian
If you just want to know how long a program took to run: time, man, man time!

@Dorian如果你只是想知道一个程序运行多长时间：时间，男人，男人的时间！

Trivial example:

琐碎的例子：

jonathan@Odin:~$ time sleep 1

real    0m1.001s
user    0m0.000s
sys     0m0.000s

OK, it doesn't give the result in seconds, but you can make it do so with a format string, or more simply with the POSIX compliance option:

好的，它不会在几秒钟内给出结果，但您可以使用格式字符串，或者更简单地使用POSIX合规性选项：

jonathan@Odin:~$ time -p sleep 20
real 20.00
user 0.00
sys 0.00

#1