Logstash grok filter apache pattern

Question

这是一个示例 Apache Tomcat 日志：

portal.portal.some.thing.int:8443 13.233.220.113 - - [09/Sep/2019:00:08:02 +0200] "GET /en/search-results?p_p_id=portal201_WAR_portal201_INSTANCE_q8EzsBteHybf&p_p_lifecycle=1&p_p_state=normal&queryText=Poll&facet.collection=AΜLex%2CAMsom%2CAMss%2WebPage%2SummariesOfSomething&startRow=1&resultsPerPage=10&SEARCH_TYPE=SIMPLE HTTP/1.1" 230 334734 6261 - - 35S64857F6860FDFC0F60B5B47A97E18
10.235.350.103 94.62.15.157, 10.435.230.101,10.134.046.2

我想捕获以下变量

09/九月/2019：00：08：02 +0200

/en/search-results？p_p_id=portal2....

35S64857F6860FDFC0F60B5B47A97E18

你能帮我吗？我只想索引那些并删除其他的，这可能吗？谢谢

Answer 1

使用以下 grok 模式：

%{GREEDYDATA:field1} %{IP:ip1} - - [%{GREEDYDATA:date}] "%{WORD:method} %{GREEDYDATA:request}" %{WORD:numbers} %{WORD:numbers} %{WORD:numbers} - - %{WORD:last_parameter}

输入：

portal.portal.some.thing.int:8443 13.233.220.113 - - [09/Sep/2019:00:08:02 +0200] "GET /en/search-results?p_p_id=portal201_WAR_portal201_INSTANCE_q8EzsBteHybf&p_p_lifecycle=1&p_p_state=normal&queryText=Poll&facet.collection=AΜLex%2CAMsom%2CAMss%2WebPage%2SummariesOfSomething&startRow=1&resultsPerPage=10&SEARCH_TYPE=SIMPLE HTTP/1.1" 230 334734 6261 - - 35S64857F6860FDFC0F60B5B47A97E18
10.235.350.103 94.62.15.157, 10.435.230.101,10.134.046.2

输出：

{
"field1": [
[
"portal.portal.some.thing.int:8443"
]
],
"ip1": [
[
"13.233.220.113"
]
],
"IPV6": [
[
null
]
],
"IPV4": [
[
"13.233.220.113"
]
],
"date": [
[
"09/Sep/2019:00:08:02 +0200"
]
],
"method": [
[
"GET"
]
],
"request": [
[
"/en/search-results?p_p_id=portal201_WAR_portal201_INSTANCE_q8EzsBteHybf&p_p_lifecycle=1&p_p_state=normal&queryText=Poll&facet.collection=AΜLex%2CAMsom%2CAMss%2WebPage%2SummariesOfSomething&startRow=1&resultsPerPage=10&SEARCH_TYPE=SIMPLE HTTP/1.1"
]
],
"numbers": [
[
"230",
"334734",
"6261"
]
],
"last_parameter": [
[
"35S64857F6860FDFC0F60B5B47A97E18"
]
]
}

所需的字段包括：

• 日期
• 请求
• last_parameter

您可以使用突变过滤器中的删除字段删除其他字段。