Bash：计算两个时间戳之间经过的时间

Question

我编写了一个脚本，当某个值不在给定范围内时会通知我。“超出范围”的所有值都记录在一组每日文件中。

每行都以专有的反向方式加时间戳：yyyymmddHHMMSS

现在，我想改进脚本，并在给定超出范围值的上次通知后至少过去 60 分钟时接收通知。

我已经解决了以相反顺序打印日志的问题：

for i in $(ls -t /var/log/logfolder/*); do zcat $i|tac|grep \!\!\!|grep --color KEYFORVALUE; done

结果是：

...
20170817041001 - WARNING: KEYFORVALUE=252.36 is not between 225 and 245 (!!!)
20170817040001 - WARNING: KEYFORVALUE=254.35 is not between 225 and 245 (!!!)
20170817035001 - WARNING: KEYFORVALUE=254.55 is not between 225 and 245 (!!!)
20170817034001 - WARNING: KEYFORVALUE=254.58 is not between 225 and 245 (!!!)
20170817033001 - WARNING: KEYFORVALUE=255.32 is not between 225 and 245 (!!!)
20170817032001 - WARNING: KEYFORVALUE=254.99 is not between 225 and 245 (!!!)
20170817031001 - WARNING: KEYFORVALUE=255.95 is not between 225 and 245 (!!!)
20170817030001 - WARNING: KEYFORVALUE=255.43 is not between 225 and 245 (!!!)
20170817025001 - WARNING: KEYFORVALUE=255.26 is not between 225 and 245 (!!!)
20170817024001 - WARNING: KEYFORVALUE=255.42 is not between 225 and 245 (!!!)
20170817012001 - WARNING: KEYFORVALUE=252.04 is not between 225 and 245 (!!!)
...

无论如何，我一直在计算其中两个时间戳之间的秒数，例如：

20170817040001
20160312000101

我应该怎么做才能计算两个时间戳之间经过的时间？

Answer 1

这将为您提供以秒为单位的日期（自 UNIX 时代以来）

date --date '2017-08-17 04:00:01' +%s    # "1502938801"

这将为您提供几秒钟内的可读字符串形式的日期

date --date '@1502938801'    # "17 Aug 2017 04:00:01"

所以所有需要的是将您的日期/时间戳转换为 GNUdate可以理解的格式，使用数学来确定差异，并输出结果

datetime1=20170817040001
datetime2=20160312000101

# bash string manipulation
datestamp1="${datetime1:0:4}-${datetime1:4:2}-${datetime1:6:2} ${datetime1:8:2}:${datetime1:10:2}:${datetime1:12:2}"
datestamp2="${datetime2:0:4}-${datetime2:4:2}-${datetime2:6:2} ${datetime2:8:2}:${datetime2:10:2}:${datetime2:12:2}"

# otherwise use sed
# datestamp1=$(echo "$datetime1" | sed -nr 's/(....)(..)(..)(..)(..)(..)/\1-\2-\3 \4:\5:\6/p')
# datestamp2=$(echo "$datetime2" | sed -nr 's/(....)(..)(..)(..)(..)(..)/\1-\2-\3 \4:\5:\6/p')

seconds1=$(date --date "$datestamp1" +%s)
seconds2=$(date --date "$datestamp2" +%s)

delta=$((seconds1 - seconds2))
echo "$delta seconds"    # "45197940 seconds"

我们没有在这里提供时区信息，所以它假设本地时区。您从日期时间开始的秒值可能与我的不同。（如果您的值是 UTC，那么您可以使用date --utc.）

Answer 2

这很容易使用包中提供的datediff命令dateutils。

ddiff -i '%Y%m%d%H%M%S' 20170817040001 20160312000101