grok filter（regex）在方括号内提取字符串

Question

我的应用程序日志条目如下：

2015-06-24 14:03:16.7288  Sent request message [649b85fa-bfa0-4cb4-8c38-1aeacd1cbf74] <Request>sometext</Request>

2015-06-24 14:38:05.2460  Received response message [649b85fa-bfa0-4cb4-8c38-1aeacd1cbf74] <Response>sometext</Response>

我正在使用logstash grok过滤器来使用方括号提取xml内容和客户端令牌。

grok {  
    match => ["message", "(?<content>(<Request(.)*?</Request>))"]   
    match => ["message", "(?<clienttoken>(Sent request message \[(.)*?\]))"]
    add_tag => "Request"
    break_on_match => false
    tag_on_failure => [ ]
}

grok {  
    match => ["message", "(?<content>(<Response(.)*?</Response>))"] 
    match => ["message", "(?<clienttoken>(Received response message \[(.)*?\]))"]
    add_tag => "Response"
    break_on_match => false
    tag_on_failure => [ ]
}

现在结果如下所示

对于第一个日志行：

Content =  <Request>sometext</Request>
clienttoken = Sent request message [649b85fa-bfa0-4cb4-8c38-1aeacd1cbf74]

对于第二个日志行：

Content = <Response>sometext</Response>
clienttoken = Received response message [649b85fa-bfa0-4cb4-8c38-1aeacd1cbf74]

但我希望结果是这样的：

Content = <Request>sometext</Request>
clienttoken = 649b85fa-bfa0-4cb4-8c38-1aeacd1cbf74

请让我知道如何只提取方括号内的字符串而没有模式中的所有匹配字符串。

Answer 1

您可以使用lookbehind和lookahead断言。

(?<=Sent request message \[).*?(?=\])

同样为响应消息做同样的事情。