问题如标题,想知道有没有什么方法可以实现,我使用了import,发现老是提示说找不到定义的Bean.
quartz.xml:
[code="java"]<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">
<bean id="trigger" class="org.springframework.scheduling.quartz.CronTriggerBean">
<property name="jobDetail">
<ref local="job" />
</property>
<property name="cronExpression" value="0/5 * * * * ?" />
</bean>
<bean id="trigger2" class="org.springframework.scheduling.quartz.CronTriggerBean">
<property name="jobDetail">
<ref local="job" />
</property>
<property name="cronExpression" value="0/10 * * * * ?" />
</bean>
<bean id="schdulerFactory"
class="org.springframework.scheduling.quartz.SchedulerFactoryBean">
<property name="triggers">
<list>
<ref local="trigger" />
<ref local="trigger2" />
</list>
</property>
</bean>
[/code]
message.xml:
[code="java"]<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">
test.MessageJob
Another Xml File!
[/code]
main.java:
[code="java"]
package test;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.FileSystemXmlApplicationContext;
public class main {
/**
* @param args
*/
public static void main(String[] args) {
ApplicationContext ctx = new FileSystemXmlApplicationContext("./bin/quartz.xml");
}
}
[/code]
MessageJob.java:
[code="java"]package test;
import java.util.Map;
import org.quartz.Job;
import org.quartz.JobExecutionContext;
import org.quartz.JobExecutionException;
public class MessageJob implements Job {
public void execute(JobExecutionContext context) throws JobExecutionException {
Map properties = context.getJobDetail().getJobDataMap();
System.out.println("Current Fire Time: " + context.getFireTime());
System.out.println(properties.get("message"));
}
}[/code]