有关 定时触发器 quartz corn表达式 0 0/10 0-23 * * ? 0-23点间 每10分钟 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 能不能把这个定时触发器的流程告诉我一下啊,这个表达式是没问题了.在servlet的init()方法里启动了该触发器JobDetail job = new JobDetail("job", "group", SendJob.class);// 启动一个定时器SchedulerFactory sf = new StdSchedulerFactory();sched = sf.getScheduler();//得到定时器表达式String cronExpression = config.getInitParameter("cronExpression");CronTrigger trigger = new CronTrigger("trigger","group","job","group",cronExpression);sched.addJob(job, true);sched.scheduleJob(trigger); sched.start();sendJob 是个发送短信的业务类 ,但是在servlet的service里调用了发送短信的业务方法,我老觉得那个定时器和这个发送短信的业务方法没有什么关联 ,还请各位给解释一下 谢谢了 ... sendJob要实现Job接口下面是个完整的演示程序package test;import java.text.ParseException;import java.util.Map;import org.quartz.CronTrigger;import org.quartz.JobDetail;import org.quartz.Scheduler;import org.quartz.SchedulerException;import org.quartz.Trigger;import org.quartz.impl.StdSchedulerFactory;public class CronTriggerExample { public static void main(String[] args) throws SchedulerException, ParseException { Scheduler scheduler=new StdSchedulerFactory().getScheduler(); scheduler.start(); JobDetail jobDetail=new JobDetail("messageJob",Scheduler.DEFAULT_GROUP,MessageJob.class); Map map=jobDetail.getJobDataMap(); map.put("message","This is a message from Quartz"); String cronExpression="* 0/10 0-23 * * ?"; Trigger trigger=new CronTrigger("cronTrigger",Scheduler.DEFAULT_GROUP,cronExpression); scheduler.scheduleJob(jobDetail,trigger); }}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("Previous Fire Time: "+context.getPreviousFireTime()); }} activeMQ 请求AjaxServlet 返回500 我想改变.properties里属性的值,但是为啥还是原来的值呢,求高手解答啊 jpa(hibernate3.5.1)延时加载 从Struts2的action中获得指定的值显示在<s:select/>上,有点特殊 jfreechart在webwork下部署,webwork无法工作 Tiles调试,eclipse代码格式化,快捷键问题 struts的关于session的问题 Hibernate 出现这种错误,是咋回事? IDEA和Tomcat下目录的问题! 请教一个问题,谢谢大家 广州求助 javamail如何实现邮件的自动回复
在servlet的init()方法里启动了该触发器
JobDetail job = new JobDetail("job", "group", SendJob.class);// 启动一个定时器
SchedulerFactory sf = new StdSchedulerFactory();
sched = sf.getScheduler();//得到定时器表达式
String cronExpression = config.getInitParameter("cronExpression");
CronTrigger trigger = new CronTrigger("trigger","group","job","group",cronExpression);sched.addJob(job, true);
sched.scheduleJob(trigger);
sched.start();
sendJob 是个发送短信的业务类 ,但是在servlet的service里调用了发送短信的业务方法,我老觉得那个定时器和这个发送短信的业务方法没有什么关联 ,还请各位给解释一下 谢谢了 ...
下面是个完整的演示程序
package test;import java.text.ParseException;
import java.util.Map;import org.quartz.CronTrigger;
import org.quartz.JobDetail;
import org.quartz.Scheduler;
import org.quartz.SchedulerException;
import org.quartz.Trigger;
import org.quartz.impl.StdSchedulerFactory;public class CronTriggerExample { public static void main(String[] args) throws SchedulerException, ParseException {
Scheduler scheduler=new StdSchedulerFactory().getScheduler();
scheduler.start();
JobDetail jobDetail=new JobDetail("messageJob",Scheduler.DEFAULT_GROUP,MessageJob.class);
Map map=jobDetail.getJobDataMap();
map.put("message","This is a message from Quartz");
String cronExpression="* 0/10 0-23 * * ?";
Trigger trigger=new CronTrigger("cronTrigger",Scheduler.DEFAULT_GROUP,cronExpression);
scheduler.scheduleJob(jobDetail,trigger);
}}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("Previous Fire Time: "+context.getPreviousFireTime());
}}