计算unix日志文件中两个时间段之间的行数 [英] Count number of lines between two time period in log file in unix

问题描述

Log file:

<l:event dateTime="2014-02-12 08:05:37.950"..........
<l:event dateTime="2014-02-12 08:08:77.980"..........
.
.
.
<l:event dateTime="2014-02-12 10:25:39.550"..........

我想数一下08:00到10:30之间的数行，怎么弄.?注意:日志文件可能有也可能没有准确时间为 08:00 或 10:30 的条目

I want to count the number lines between 08:00 to 10:30, how to get it. ? Note: Log file may or may not have the entry with exact time of 08:00 or 10:30

推荐答案

你可以使用 perl(无论如何都不是 perl 大师，所以可能比必要的更复杂)

You could use perl (not a perl guru by any stretch, so may be more complex than necessary)

perl -n -e 'BEGIN {$cnt=0} END { print $cnt."\n"} /dateTime="\d{4}-\d{2}-\d{2} (\d{2}:\d{2}:\d{2})/ && $1 ge "08:00:00" && $1 lt "10:00:00" && $cnt++' < log.txt

...或为了可读性；

perl -n                  -- runs the script for each line in the input file

BEGIN { $cnt=0 }         -- start by setting $cnt to 0
END { print $cnt."\n"}   -- when all is done, print $cnt
/dateTime="\d{4}-\d{2}-\d{2} (\d{2}:\d{2}:\d{2})/
                         -- match for time format, keeping the time in the group
$1 ge "08:00:00"         -- check if the time is greater or equal to 08:00:00
$1 lt "10:30:00"         -- check if time is less than 10:30:00 
$cnt++                   -- if all matches are ok, increase cnt

从评论中编辑；

/dateTime="\d{4}-\d{2}-\d{2} (\d{2}:\d{2}:\d{2})/

...基本上是一个正则表达式，它将行与您提供的日期时间字段格式相匹配(4 位数字 + - + 2 位数字 + - + ...) 并将时间部分(括号中的部分)提取到 $1 中以与限制进行比较.它应该适用于一天内的任何时间跨度，因此 10:25-10:35 应该可以正常工作.

...basically is a regex that matches the rows with the datetime field format you're giving (4 digits + - + 2 digits + - + ...) and extracts the time part (the parenthesized part) into $1 for comparison with the limits. It should work for any time span within a single day, so 10:25-10:35 should work just fine.

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