linux - grep 选择行

Question

我想从日志文件中选择最近7天的记录也包含“OK LOGIN”

 Sat Sep  2 03:32:13 2012 [pid 12461] CONNECT: Client "66.249.68.236"
 Sat Sep  2 03:32:13 2012 [pid 12460] [ftp] OK LOGIN: Client "66.249.68.236", anon     password "gxxglxxxxt@google.com"
 Sat Sep  2 03:32:14 2012 [pid 12462] [ftp] OK DOWNLOAD: Client "66.249.68.236",   "/pub/10.5524/100001_101000/100022/readme.txt", 451 bytes, 1.39Kbyte/sec
 Sat Sep  2 03:32:22 2012 [pid 12677] CONNECT: Client "66.249.68.236"
 Sat Sep  2 03:32:23 2012 [pid 12676] [ftp] OK LOGIN: Client "66.249.68.236", anon password "xxxxxbot@google.com"

我使用这个命令，但它不起作用以及如何编写如果我想要 date= [当前日期 - 7 天]

 cat vsftp.log | grep 'OK LOGIN' |egrep "Sep [1-6]"

Answer 1

你可以use sed to select range of lines and generate the regex using date :

#!/bin/bash

function md()
{
    date +'%b %e' --date "$@"
}

start=`md 'last week'`
end=`md 'now'`
grep 'OK LOGIN' | sed -n "/$start/,/$end/p"

每天应该至少有一次登录。正则表达式是允许的，可能存在误报。