如何根据时间条件在bash中拆分日志文件 [英] How to split log file in bash based on time condition

问题描述

我有一个带有时间戳的简单日志文件，以毫秒为单位:

I have simple log file with timestamp in milliseconds like this one:

[02/03/2020 07:53:58.859000] 5
[02/03/2020 07:53:58.935300] 2
[02/03/2020 10:04:50.355600] 0
[02/03/2020 10:04:51.028900] 1
[02/03/2020 10:38:13.468200] 6

我想获得更好的可读性，因此+ -2秒以这样的破折号分隔日志:

And I want to achieve better readability so +-2seconds separate logs by dashes like this one:

[02/03/2020 07:53:58.859000] 5
[02/03/2020 07:53:58.935300] 2
------------------------------
[02/03/2020 10:04:50.355600] 0
[02/03/2020 10:04:51.028900] 1
------------------------------
[02/03/2020 10:38:13.468200] 6

如何通过bash脚本中的简单循环来实现它?到目前为止，我已经弄清楚如何从字符串 NEW_VALUE1 ="$(date -d" $ VALUE 2 seconds"+'％d/％m/％Y％H:％M:％S'格式化和修改日期)"，但没有运气将其实现为功能性结果.

How to achieve it by simple loop in bash script? So far I figured out how to format and modify date from string NEW_VALUE1="$(date -d "$VALUE 2 seconds" +'%d/%m/%Y %H:%M:%S')" but with no luck to implement it to functional result.

推荐答案

使用GNU awk :

awk -F'[[/:. ]' '
  { t=mktime($4" "$3" "$2" "$5" "$6" "$7) }
  NR>1 && t>tlast+2 { print "------------------------------" }; 1
  { tlast=t }
' file

• 使用 [，/，: ..和空格字符作为字段分隔符，并创建时间戳记每行 t .
• 如果不是第一行并且 t> ;，则打印分隔线.tlast + 2 .
• 打印当前行.
• 将 t 的值分配给 tlast .
• Use [, /, : . and the space character as field separator characters and create a timestamp t for each line.
• Print a separator line if this is not the first line and if t > tlast + 2.
• Print the current line.
• Assign value of t to tlast.

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