Question

我有一个HAProxy日志文件，其内容与此类似：

Feb 28 11:16:10 localhost haproxy[20072]: 88.88.88.88:6152 [28/Feb/2017:11:16:01.220] frontend backend_srvs/srv1 9063/0/0/39/9102 200 694 - - --VN 9984/5492/191/44/0 0/0 {Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.87 Safari/537.36|http://subdomain.domain.com/location1} "GET /location1 HTTP/1.1"
Feb 28 11:16:10 localhost haproxy[20072]: 88.88.88.88:6152 [28/Feb/2017:11:16:10.322] frontend backend_srvs/srv1 513/0/0/124/637 200 14381 - - --VN 9970/5491/223/55/0 0/0 {Mozilla/5.0 AppleWebKit/537.36 Chrome/56.0.2924.87 Safari/537.36|http://subdomain.domain.com/location2} "GET /location2 HTTP/1.1"
Feb 28 11:16:13 localhost haproxy[20072]: 88.88.88.88:6152 [28/Feb/2017:11:16:10.960] frontend backend_srvs/srv1 2245/0/0/3/2248 200 7448 - - --VN 9998/5522/263/54/0 0/0 {another user agent with fewer columns|http://subdomain.domain.com/location3} "GET /location3 HTTP/1.1"
Feb 28 11:16:13 localhost haproxy[20072]: 88.88.88.88:6152 [28/Feb/2017:11:16:10.960] frontend backend_srvs/srv1 2245/0/0/3/2248 200 7448 - - --VN 9998/5522/263/54/0 0/0 {Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.87 Safari/537.36|} "GET /another_location HTTP/1.1"

我想提取一些字段以获得以下输出：

 Field 1             Field 2           Field 3         Field 4         Field 5         Field 6
Date/time       HTTP status code     HTTP Method       Request      HTTP version    Referer URL

基本上，在这种特殊情况下，输出应为：

Feb 28 11:16:10  200 GET /location1 HTTP/1.1    http://subdomain.domain.com/location1
Feb 28 11:16:10  200 GET /location2 HTTP/1.1    http://subdomain.domain.com/location2
Feb 28 11:16:13  200 GET /location3 HTTP/1.1    http://subdomain.domain.com/location3
Feb 28 11:16:13  200 GET /another_location HTTP/1.1

此处唯一的问题是提取Referer URL，该URL位于大括号和用户代理之间，并且它们由管道分隔。此外，用户代理具有可变数量的字段。

我能想到的唯一解决方案是分别提取引用URL，然后将列粘贴在一起：

requests_temp=`grep -F " 88.88.88.88:" /root/file.log | tr -d '"'`
requests=`echo "${requests_temp}" | awk '{print $1" "$2" "$3"  "$11, $(NF-2), $(NF-1), $NF}' > /tmp/requests_tmp`
referer_url=`echo "${requests_temp}" | awk 'NR > 1 {print $1}' RS='{' FS='}' | awk -F'|' '{ print $2 }' > /tmp/referer_url_tmp`

paste /tmp/abuse_requests_tmp /tmp/referer_url_tmp

但我不喜欢这种方法。有没有其他方法可以使用只有一个awk线来做到这一点？也许将referer url列分配给awk中的变量，然后用它来创建相同的输出？

Answer 1

您可以使用awk：

一次完成所有操作

awk '$6 ~ /88\.88\.88\.88:[0-9]+/{
   split($0,a,/[{}]/)
   $0=a[1] OFS a[3]
   split(a[2],b,"|")
   print $1,$2,$3,$11,substr($18,2),$19,substr($20,1,length($20)-1),b[2]
}' file.log

第一个split将行的可变部分（包含在{...}之间）拆分为数组a。

重建该行以获得固定数量的字段$0=a[1] OFS a[3]

第二个split允许根据|字符从变量中提取网址。

最后，print显示了所有需要的元素。请注意substr用于删除"。

Answer 2

尝试以下解决方案 -

awk '/88.88.88.88/ {gsub(/"/,"",$0);split($(NF-3),a,"|"); {print $1,$2,$3,$11, $(NF-2), $(NF-1), $NF, substr(a[2],1,(length(a[2])-1))}}' a
Feb 28 11:16:10 200 GET /location1 HTTP/1.1 http://subdomain.domain.com/location1
Feb 28 11:16:10 200 GET /location2 HTTP/1.1 http://subdomain.domain.com/location2
Feb 28 11:16:13 200 GET /location3 HTTP/1.1 http://subdomain.domain.com/location3
Feb 28 11:16:13 200 GET /another_location HTTP/1.1

将另一列添加到awk输出

2 个答案: