Question

我有access.log文件，其中包含超过1000个X-Forwarded-For日志条目，如下所示

142.245.59.16, 67.69.175.224, 69.31.97.126 - - [22/Sep/2015:20:00:02 -0400] "GET /company-information/cs/null?path=%
157.55.39.76, 184.27.179.176, 165.254.1.175 - - [22/Sep/2015:20:00:05 -0400] "GET /metricstream/--ID__100325--/free-co-profile.xhtml
10.70.33.32 - - [22/Sep/2015:20:00:22 -0400] "GET /autodiscover/autodiscover.xml
172.30.152.90, 198.178.234.30, 184.27.120.46, 69.31.97.126 - - [22/Sep/2015:20:03:37 -0400] "GET /company-information/cs/null?path

使用此日志条目，我必须grep并将它们解压缩到access_log.txt文件，如下面的输出

142.245.59.16 - - [22/Sep/2015:20:00:02 -0400] "GET /company-information/cs/null?path=%
157.55.39.76 - - [22/Sep/2015:20:00:05 -0400] "GET /metricstream/--ID__100325--/free-co-profile.xhtml
10.70.33.32 - - [22/Sep/2015:20:00:22 -0400] "GET /autodiscover/autodiscover.xml
172.30.152.90 - - [22/Sep/2015:20:03:37 -0400] "GET /company-information/, csnull ?path

这是留下第一个ip，并删除以下两个或更多的ips，我也厌倦了REGEX：/\, .*?\ -/g但我不知道如何在unix sed命令中应用它。请使用Unix命令

帮助解决这个问题

Answer 1

您可以使用此sed命令：

sed 's/, [^-]*- -/ - -/' file.log
142.245.59.16 - - [22/Sep/2015:20:00:02 -0400] "GET /company-information/cs/null?path=%
157.55.39.76 - - [22/Sep/2015:20:00:05 -0400] "GET /metricstream/--ID__100325--/free-co-profile.xhtml
10.70.33.32 - - [22/Sep/2015:20:00:22 -0400] "GET /autodiscover/autodiscover.xml
172.30.152.90 - - [22/Sep/2015:20:03:37 -0400] "GET /company-information/cs/null?path

Answer 2

那样： sed＆＃39; s / \，。* - / - / g＆＃39; ./access.log

使用unix删除日志文件中的以下ips

2 个答案: