解析某些时间戳之间的文件内容 [英] Parse contents of a file between certain timestamps

问题描述

我有一个包含了大量的信息的日志文件，我想只解析该文件落入过去24小时内的内容

I have a log file that contains a lot of information and I would like to only parse the contents of that file which fall within the last 24 hours

文件中的每一行与时间戳开头，如 1月31日13时13分02秒，然后有一个日志信息。

Each line in the file begins with a timestamp such as Jan 31 13:13:02 and then has a log message.

我目前有发现开始和结束时间这样的批处理文件

I currently have a batch file that finds the start and end time like this

start=$(date +"%b  %d %H:%M:%S" --date="-1 day")
end=$(date +"%b  %d %H:%M:%S")

我当时希望利用这些时间用的grep -cdata_to_find一起查找某一日志消息出现的次数，这样我可以然后采取行动关于这个版本。

I was then hoping to use these times along with a grep -c "data_to_find" to find the number of occurrences of a certain log message so that I can then act on this later.

总之，我怎样才能兼顾次，然后用grep一个字符串的出现次数的内容中说文件？

In short, How can I take into account the times and then grep the content for the number of occurrences of a string within said file?

我在Linux系统上，并有与使用SED，AWK，任何解决方案没有问题，GREP等。

I am on a linux system and have no issue with any solution that uses SED, AWK, GREP etc.

推荐答案

无需编写shell脚本（特别是如果它不排序）没那么简单。

Not so simple without writing a shell script (especially if it's not sorted).

我会尝试这样的事情来获得在1天前，现在的grep -c 管不管你想输出的所有行（插值如果需要），然后。下面的说明假定日期格式是一样的东西 1月31日13时13分02秒（月和日之间的2个空格）

I would try something like this to get all the lines between 1 day ago and now (interpolate as needed), and then grep -c pipe whatever you want from output. Note below assumes date format is something like Jan 31 13:13:02 (2 spaces between Month and Day)

#!/bin/bash
yest=$(date -d "1 days ago" '+%s')
today=$(date '+%s')

while read -r line; do
  date=
  [[ $line =~ ^[[:alpha:]]+[[:blank:]][[:blank:]][0-9]+[[:blank:]][0-9]+':'[0-9]+':'[0-9]+ ]] && date="${BASH_REMATCH[0]}"
  [[ -n $date ]] && date=$(date -d "$date" '+%s')
  [[ -n $date && $date -ge $yest && $date -le $today ]] && echo "$line"
done < logfile

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