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[英]How to gather IP and User Agent info and uniq them base IP address from nginx access log with AWK?
[英]How to gather IP and User Agent info from web log with AWK?
我有一個日志文件,包含如下文字:
66.249.74.18 - - [21/Apr/2013:05:55:33 +0000] 200 "GET /1.jpg HTTP/1.1" 7691 "-" "Googlebot-Image/1.0" "-" 220.181.108.96 - - [21/Apr/2013:05:55:33 +0000] 200 "GET /1.html HTTP/1.1" 17722 "-" "Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)" "-"
我想將所有ip和用戶代理信息收集到一個文件中:
66.249.74.18 "Googlebot-Image/1.0" 220.181.108.96 "Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)"
我怎么能用awk做到這一點?
我知道awk '{print $1}'
可以列出所有ips和awk -F\\" '{print $6}'
可以列出所有用戶代理,但我不知道如何將它們組合成輸出。
awk '{print $1,$6}' FPAT='(^| )[0-9.]+|"[^"]*"'
[0-9.]+
或"[^"]*"
不使用GNU擴展的可移植方法:
awk '{printf "%s ",$1;for(i=12;i<NF;i++)printf "%s ",$i;printf "\n"}' file
awk -F' - |\\"' '{print $1, $7}' temp1
輸出:
66.249.74.18 Googlebot-Image/1.0
220.181.108.96 Mozilla/5.0 (compatible;Baiduspider/2.0;+http://www.baidu.com/search/spider.html)
temp1文件:
66.249.74.18 - - [21/Apr/2013:05:55:33 +0000] 200 "GET /1.jpg HTTP/1.1" 7691 "-" "Googlebot-Image/1.0" "-"
220.181.108.96 - - [21/Apr/2013:05:55:33 +0000] 200 "GET /1.html HTTP/1.1" 17722 "-" "Mozilla/5.0 (compatible; Baiduspider/2.0; +http://www.baidu.com/search/spider.html)" "-"
使用perl
:
perl -nle '/^((?:\d+\.?){4})(?:.+?"){4}\s+(".*?")/ && print "$1 $2"' access_log
訣竅在於計算不是雙引號+雙引號的字符 : (?:.+?"){4}
。這是正則表達式的直觀描述: https : //regex101.com/r/xP0kF4/4
正則表達式比以前的答案更復雜,但我們可以輕松地解析其他屬性。
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