a1.sources = r1
a1.sinks = k1 k2
a1.channels = c1 c2 # Describe/configure the source
a1.sources.r1.type = netcat
a1.sources.r1.bind = localhost
a1.sources.r1.port = 44444 # Describe the sink
a1.sinks.k1.type =file_roll
a1.sinks.k1.sink.directory=/home/chenyun/data/flume/file_sinke1 a1.sinks.k2.type =file_roll
a1.sinks.k2.sink.directory=/home/chenyun/data/flume/file_sinke2 # Use a channel which buffers events in memory
a1.channels.c1.type = memory
a1.channels.c1.capacity = 1000
a1.channels.c1.transactionCapacity = 100 a1.channels.c2.type = memory
a1.channels.c2.capacity = 1000
a1.channels.c2.transactionCapacity = 100 # Bind the source and sink to the channel
a1.sources.r1.channels = c1 c2
a1.sources.r1.interceptors = i1
a1.sources.r1.interceptors.i1.type = regex_extractor
a1.sources.r1.interceptors.i1.regex = (\\w+):(\\w+):(\\w+)
a1.sources.r1.interceptors.i1.serializers = s1 s2 s3
a1.sources.r1.interceptors.i1.serializers.s1.name = one
a1.sources.r1.interceptors.i1.serializers.s2.name = two
a1.sources.r1.interceptors.i1.serializers.s3.name = three
a1.sources.r1.selector.type = multiplexing
a1.sources.r1.selector.header = one
a1.sources.r1.selector.mapping.hadoop = c1
a1.sources.r1.selector.default = c2
a1.sinks.k1.channel=c1
a1.sinks.k2.channel = c2

Hadoop实战-Flume之Source multiplexing(十五)的更多相关文章

  1. Hadoop实战-Flume之Source replicating(十四)

    a1.sources = r1 a1.sinks = k1 k2 a1.channels = c1 c2 # Describe/configure the source a1.sources.r1.t ...

  2. Hadoop实战-Flume之Source regex_extractor(十二)

    a1.sources = r1 a1.sinks = k1 a1.channels = c1 # Describe/configure the source a1.sources.r1.type = ...

  3. Hadoop实战-Flume之Sink Failover(十六)

    a1.sources = r1 a1.sinks = k1 k2 a1.channels = c1 # Describe/configure the source a1.sources.r1.type ...

  4. Hadoop实战-Flume之Hdfs Sink(十)

    a1.sources = r1 a1.sinks = k1 a1.channels = c1 # Describe/configure the source a1.sources.r1.type = ...

  5. Hadoop实战-Flume之自定义Sink(十九)

    import java.io.File; import java.io.FileNotFoundException; import java.io.FileOutputStream; import j ...

  6. Hadoop实战-Flume之Source regex_filter(十三)

    a1.sources = r1 a1.sinks = k1 a1.channels = c1 # Describe/configure the source a1.sources.r1.type = ...

  7. Hadoop实战-Flume之Source interceptor(十一)(2017-05-16 22:40)

    a1.sources = r1 a1.sinks = k1 a1.channels = c1 # Describe/configure the source a1.sources.r1.type = ...

  8. .NET Core实战项目之CMS 第十五章 各层联动工作实现增删改查业务

    连着两天更新叙述性的文章大家可别以为我转行了!哈哈!今天就继续讲讲我们的.NET Core实战项目之CMS系统的教程吧!这个系列教程拖得太久了,所以今天我就以菜单部分的增删改查为例来讲述下我的项目分层 ...

  9. Hadoop实战-Flume之自定义Source(十八)

    import java.nio.charset.Charset; import java.util.HashMap; import java.util.Random; import org.apach ...

随机推荐

  1. luogu P1608 路径统计

    题目描述 “RP餐厅”的员工素质就是不一般,在齐刷刷的算出同一个电话号码之后,就准备让HZH,TZY去送快餐了,他们将自己居住的城市画了一张地图,已知在他们的地图上,有N个地方,而且他们目前处在标注为 ...

  2. JSP介绍与语法-java之JSP学习第一天(非原创)

    文章大纲 一.JSP 简介二.JSP 生命周期三.JSP 语法四.学习资料下载五.参考文章   一.JSP 简介 1. 什么是Java Server Pages? JSP全称Java Server P ...

  3. Zabbix 企业Nginx监控

    Zabbix监控Nginx状态 1 修改Nginx配置文件,开启Nginx监控 location /nginx_status { stub_status on; access_log off; all ...

  4. Jenkins错误“to depth infinity with ignoreexternals:true”问题解决

    试下以下解决方法: 1.可能是SVN插件版本过低导致,升级SVN插件. 2.可能是构建时自己手动修改了代码,而SVN检出时无法覆盖导致的错误,可以先删除jenkins检出的代码,然后再检出一次去构建. ...

  5. Difference between a Hard Link and Soft (Symbolic) Link

    Within the Unix/Linux file system, linking lets you create file shortcuts to link one or more files. ...

  6. 受检查异常要求try catch,new对象时,就会在堆中创建内存空间,创建的空间包括各个成员变量类型所占用的内存空间

    ,new对象时,就会在堆中创建内存空间,创建的空间包括各个成员变量类型所占用的内存空间

  7. 【java】hash一致性算法的实现区别【标题暂定】

    下面两个都是在生成sign签名时候用到的方式,有什么区别? 第一种: import org.apache.commons.codec.digest.DigestUtils; String sign = ...

  8. JAVAWEB开发之JSP、EL、及会话技术(Cookie和Session)的使用详解

    Servlet的缺点 开发人员要十分熟悉JAVA 不利于页面调试和维护(修改,重新编译) 很难利用网页设计工具进行页面设计(HTML内容导入到servlet中,用PrintWriter的对象进行输出) ...

  9. 【好】Paxos以及分布式一致性的学习

    Paxos,一言以蔽之,我们需要一种提交协议来确保分布式系统中的全局操作即使是在发生故障的情况下也能保证正确性. 跟拜占庭将军问题是不同的问题,虽然拜占庭也是Lamport提出的.拜占庭里面有叛徒,有 ...

  10. 140. Word Break II(hard)

    欢迎fork and star:Nowcoder-Repository-github 140. Word Break II 题目: Given a non-empty string s and a d ...