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使用 sed\awk 显示两个连续行之间不同的时间戳

[英]Using sed\awk to display timestamp different between two consequetive lines

 原文  2020-11-30 09:38:58  0   1    linux/ awk/ sed

问题描述

I have a log file displays lines like the following:我有一个日志文件显示如下行：

7,5580,70930902,-;2c04000.vdi-ch0: start 7,5580,70930902,-;2c04000.vdi-ch0: 开始
7,5621,70958680,-;2c04000.vdi-ch0: start 7,5621,70958680,-;2c04000.vdi-ch0: 开始
7,5662,70986453,-;2c04000.vdi-ch0: start 7,5662,70986453,-;2c04000.vdi-ch0: 开始
7,5703,71014240,-;2c04000.vdi-ch0: start 7,5703,71014240,-;2c04000.vdi-ch0: 开始

the third value after the seconad comma (70930902, 70958680...) is a timestamp.第二个逗号后的第三个值（70930902、70958680...）是时间戳。 I would like to display at the end of each line the difeerent between two consequetive timestamps.我想在每行的末尾显示两个连续时间戳之间的差异。 Can you please help me to figure out how to do that?你能帮我弄清楚怎么做吗？

I'm a little new with all of the sed\awk stuff, so any detaled answer will be appriciated.我对所有 sed\awk 的东西都有点陌生，所以任何详细的答案都会被应用。

Thanks谢谢
Rachel雷切尔

1 个解决方案

解决方案1
 0  2020-11-30 09:59:21

Finally I managed to write a command:最后我设法写了一个命令：

awk -F',' '{print NR,$0,"diff = ",a-$3}{a=$3}'

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