2012-07-29 40 views
8

如何將現有的apache日誌文件按月分成單獨的文件?如何按月分割現有的apache日誌文件?

我搜遍了網頁,但找不到任何東西。是的,我知道關於logrotate和cronolog等等。但是我找到的東西並沒有幫助我分割現有的文件。

是否有awk腳本或其他?

下面是數據的一個片段:

124.115.5.11 - - [30/May/2011:23:21:37 -0500] "GET/HTTP/1.0" 200 206492 "-" "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322;TencentTraveler)" 
58.61.164.39 - - [31/May/2011:00:36:35 -0500] "GET/HTTP/1.0" 200 206492 "-" "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322;TencentTraveler)" 
114.80.93.55 - - [31/May/2011:01:42:17 -0500] "GET/HTTP/1.0" 200 206492 "-" "Sosospider+(+http://help.soso.com/webspider.htm)" 
114.80.93.73 - - [31/May/2011:02:03:44 -0500] "GET/HTTP/1.0" 200 206492 "-" "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322;TencentTraveler)" 
123.125.71.98 - - [31/May/2011:12:33:30 -0500] "GET/HTTP/1.1" 103 24576 "-" "Baiduspider+(+http://www.baidu.com/search/spider.htm)" 
220.181.108.187 - - [31/May/2011:12:33:55 -0500] "GET/HTTP/1.1" 103 24576 "-" "Baiduspider+(+http://www.baidu.com/search/spider.htm)" 
123.125.71.117 - - [31/May/2011:13:27:56 -0500] "GET/HTTP/1.1" 103 24576 "-" "Baiduspider+(+http://www.baidu.com/search/spider.htm)" 
123.125.71.78 - - [31/May/2011:16:45:48 -0500] "GET /node/54 HTTP/1.1" 200 3219 "-" "Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)" 
124.115.1.8 - - [31/May/2011:19:59:58 -0500] "GET/HTTP/1.1" 200 206492 "-" "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1)" 
123.125.71.69 - - [31/May/2011:22:05:46 -0500] "GET/HTTP/1.1" 200 206492 "-" "Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)" 

這裏是我的解決辦法,按以下史蒂夫的答案極大地鼓舞:

一個使用awk方式:

awk 'BEGIN { 
    split("Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec ", months, " ") 
    for (a = 1; a <= 12; a++) 
     m[months[a]] = a 
} 
{ 
    year = array[3] 
    month = sprintf("%02d", m[array[2]]) 

    split($4,array,"[:/]"); 
    print > FILENAME"-"year"_"month".txt" 
}' incendiary.ws-2009 

這將輸出文件如:

incendiary.ws-2010-2010_04.txt 
incendiary.ws-2010-2010_05.txt 
incendiary.ws-2010-2010_06.txt 
incendiary.ws-2010-2010_07.txt 

而針對150 MB的日誌文件,通過chepner接受的答案了70秒上的3.4GHz的8核至強E31270,而這種方法把5秒

最初靈感:https://stackoverflow.com/a/11714105/430062

+1

的人,誰知道AWK(或東西:)可能不一定知道或訪問您試圖數據文件操縱,如果你可以提供一些輸入/輸出對來顯示你正在使用/想要脫離的話 – Levon 2012-07-29 23:57:26

+0

我已經實現了你的出色建議。 – 2012-07-30 00:08:01

回答

12

一個使用awk方式:

awk '{ split($4,array,"/"); print > array[2] ".txt" }' file.txt 

這將輸出文件,如:

May.txt 
June.txt 
July.txt 
etc 

編輯:

也許你想保持這些年來分開:

awk '{ split($4,array,"[:/]"); print > array[2] array[3] ".txt" }' file.txt 

這將輸出文件,如:

May2011.txt 
May2012.txt 
July2011.txt 
etc 
+1

我修改了你的算法,使格式FILENAME-YYYY_MM。我堅持它在我的問題。 – 2012-07-30 12:15:23