[英]Simple Apache Camel SNMP trap
我是Apache Camel的新手,正在尝试接收一个简单的SNMP陷阱。
我已经使用camel-core和org.apache.servicemix.bundles.snmp4j设置了Maven项目。
我无法找到任何SNMP示例,但是基于其他示例,我想到了这个Main类:
public class Main {
public static Processor myProcessor = new Processor() {
public void process(Exchange arg0) throws Exception {
// save to database
}
};
public static void main(String[] args) {
CamelContext context = new DefaultCamelContext();
context.addComponent("snmp", new SnmpComponent());
RouteBuilder builder = new RouteBuilder() {
public void configure() {
from("snmp:127.0.0.1:162?protocol=udp&type=TRAP").process(myProcessor);
}
};
try {
context.addRoutes(builder);
context.start();
} catch (Exception e) {
e.printStackTrace();
}
}
}
但是,当我在Eclipse中作为Java应用程序运行它时,它会在运行半秒钟后退出。 我期望它继续运行并听127.0.0.1:162 ...
任何帮助是极大的赞赏
至少拾起一个陷阱并打印到System.out的一种方法是这样的:
public class SNMPTrap {
private Main main;
public static void main(String[] args) throws Exception {
SNMPTrap snmpTrap = new SNMPTrap();
snmpTrap.boot();
}
@SuppressWarnings("deprecation")
public void boot() throws Exception {
main = new Main();
main.bind("snmp", new SnmpComponent());
main.addRouteBuilder(new MyRouteBuilder());
main.addMainListener(new Events());
System.out.println("Starting SNMPTrap. Use ctrl + c to terminate the JVM.\n");
main.run();
}
private static class MyRouteBuilder extends RouteBuilder {
@Override
public void configure() throws Exception {
from("snmp:127.0.0.1:162?protocol=udp&type=TRAP").process(myProcessor)
.bean("snmp");
}
}
public static Processor myProcessor = new Processor() {
public void process(Exchange trap) throws Exception {
System.out.println(trap.getIn().getBody(String.class));
// Save to DB or do other good stuff
}
};
public static class Events extends MainListenerSupport {
@Override
public void afterStart(MainSupport main) {
System.out.println("SNMPTrap is now started!");
}
@Override
public void beforeStop(MainSupport main) {
System.out.println("SNMPTrap is now being stopped!");
}
}
}
但是,我得到警告,现在不赞成使用Camel核心的Main。
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