bash 在bash脚本中减去两个时间戳
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/41793634/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Subtracting two timestamps in bash script
提问by Ali
I have a file that contains a set of Timestamps in format of H:M:S.MS, I can read and print all of the saved Timestamps but when I do some arithmetic operation like Subtract (say Timestamp2 - Timestamp1) it doesn't give me the answer, I do not have experience work with bash script programming.
我有一个文件，其中包含一组格式为 的时间戳H:M:S.MS，我可以读取和打印所有保存的时间戳，但是当我执行一些算术运算时，例如减去（例如Timestamp2 - Timestamp1）它没有给我答案，我没有经验使用 bash 脚本编程。
Any suggestions, recommendations will be much appreciated.
任何建议，建议将不胜感激。
Here is the screen shotof my problem.
这是我的问题的屏幕截图。
Here is an example :
这是一个例子：
Start Time = 17:53:01.166721
End Time = 17:53:01.369787
The expected result for End Time - Start Timeis either 0:00:0.203066or 0.203066.
的预期结果End Time - Start Time是0:00:0.203066或0.203066。
回答by Grisha Levit
If you want to process the date using simple command line tools, you need to convert the timestamps into some easy-to-deal-with format, like epoch-based.
如果您想使用简单的命令行工具处理日期，则需要将时间戳转换为某种易于处理的格式，例如基于 epoch 的格式。
You can do this with the datecommand, which you can wrap in a function like so:
您可以使用date命令执行此操作，您可以将其包装在一个函数中，如下所示：
timestamp() {
date '+%s%N' --date=""
}
# timestamp "12:20:45.12345" ==> 1485105645123450000
Then you can subtract them directly:
然后你可以直接减去它们：
echo $(( $(timestamp "${Stime[$i]}") - $(timestamp "${Rtime[$i]}") ))
Note that since you are not keeping track of the day but only of the time, you will have errors when, for example, the start time is 23:59:59 and the end time is 00:00:01.
请注意，由于您不是在跟踪当天，而只是在跟踪时间，例如，当开始时间为 23:59:59 和结束时间为 00:00:01 时，您将出现错误。
回答by glenn Hymanman
If you need to use bash, you need to convert the timestamp into an integer value:
如果需要使用bash，则需要将时间戳转换为整数值：
$ time="12:34:56.789"
$ IFS=":." read -r h m s ms <<<"$time"
$ echo $h $m $s $ms
12 34 56 789
$ milliseconds=$(( (h*3600 + m*60 + s)*1000 + ms ))
$ echo $milliseconds
45296789
Then you can subtract to get a number of milliseconds representing the diff.
然后您可以减去以获得表示差异的毫秒数。
Convert to seconds with some arithmetic and string construction:
使用一些算术和字符串构造转换为秒：
$ seconds="$((milliseconds / 1000)).$((milliseconds % 1000))"
$ echo $seconds
45296.789
To address gniourf_gniourf's valid point (in a clunky way):
要解决 gniourf_gniourf 的有效观点（以笨拙的方式）：
$ time="12:34:56.7"
$ IFS=":." read -r h m s ms <<<"$time"
$ ms="${ms}000"; ms=${ms:0:3}
$ echo $h $m $s $ms
12 34 56 700
# .......^^^
Also, strings that are invalid octal numbers like "08" and "09" will throw errors in bash arithmetic, so explicitly declare they are base 10
此外，像 "08" 和 "09" 这样的无效八进制数的字符串会在 bash 算术中抛出错误，因此明确声明它们是以 10 为基数的
milliseconds=$(( (10#$h*3600 + 10#$m*60 + 10#$s)*1000 + 10#$ms ))
回答by Fred
Assuming your variables are integers, you need to enclose your calculation in an arithmetic evaluation for that to work.
假设您的变量是整数，您需要将您的计算包含在算术评估中才能使其工作。
echo $(( VAR1 - VAR2 ))
Or, if you want to assign the value :
或者，如果要分配值：
VAR3=$(( VAR1 - VAR2 ))
