CronSequenceGenerator的解决方法一个月的最后一天？

Z时代
2024-01-10
分类：问答

好了，这是我要安排的任务在每月的最后一天上午10:10运行。我的cron表达式是

0 10 10 L * ?

现在的问题是CronSequenceGenerator向’L’值抛出NumberFormatException，这意味着Spring的CronSequenceGenerator不支持这种表达式，如何以任何其他方式（解决方法）来实现，我不想使用石英或做Spring的将在新版本中对此提供支持。

这是完整的堆栈跟踪：

Exception in thread "main" java.lang.NumberFormatException: For input string: "L"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:492)
    at java.lang.Integer.valueOf(Integer.java:582)
    at org.springframework.scheduling.support.CronSequenceGenerator.getRange(CronSequenceGenerator.java:324)
    at org.springframework.scheduling.support.CronSequenceGenerator.setNumberHits(CronSequenceGenerator.java:297)
    at org.springframework.scheduling.support.CronSequenceGenerator.setDays(CronSequenceGenerator.java:275)
    at org.springframework.scheduling.support.CronSequenceGenerator.setDaysOfMonth(CronSequenceGenerator.java:266)
    at org.springframework.scheduling.support.CronSequenceGenerator.parse(CronSequenceGenerator.java:239)
    at org.springframework.scheduling.support.CronSequenceGenerator.<init>(CronSequenceGenerator.java:81)
    at org.springframework.scheduling.support.CronTrigger.<init>(CronTrigger.java:54)
    at org.springframework.scheduling.support.CronTrigger.<init>(CronTrigger.java:44)
    at com.hcdc.coedp.datantar.scheduler.SchedulerUtil.start(SchedulerUtil.java:75)
    at com.hcdc.coedp.datantar.scheduler.SchedulerUtil.changeTrigger(SchedulerUtil.java:106)
    at com.hcdc.coedp.datantar.scheduler.SchedulingService.scheduleTransfer(SchedulingService.java:70)
    at com.hcdc.coedp.datantar.scheduler.Scheduler.schedule(Scheduler.java:107)
    at main.Main.main(Main.java:47)

更新：

以下是我的调度方法

 /**
    * Schedule a task {@link Task} with a specified cron expression.
    * @param task {@link Task}
    * @param cronExpression cron expression to be applied must be a vaild one.
    * @param taskName
    * @return 
    */
     public String start(Task task, String cronExpression, String taskName) {
        CronTrigger trigger = new CronTrigger(cronExpression);//line 2
        CronSequenceGenerator generator = new CronSequenceGenerator(cronExpression, TimeZone.getTimeZone("GMT+5:30"));
        List<Date> dateList = new ArrayList<>(5);
        Date currentDate = new Date();
        for (int i = 0; i < 5; i++) {
            currentDate = generator.next(currentDate);
            dateList.add((currentDate));
            System.out.println("Next Exceution times are" + currentDate);
        }
        ScheduledFuture sf = tps.schedule(task, trigger);
        //TODO Save this scheduled future with a specific task name.
        ContextHolder.schduledFutureMap.put(taskName, sf);
        return cronExpression;
    }

在第2行，当我传递指定的cron表达式时，它引发NumberFormatException。

回答：

此功能不是标准cron表达式语法。因此，Spring可能永远不会实现它。看代码，我看不到任何

外科解决方案的扩展CronSequenceGenerator。那么，为什么不使用Quartz却是一个特殊功能呢？

根据您的确切需求，您可以实现自己的Trigger。就像是：

import java.util.Date;
import org.joda.time.LocalDate;
import org.joda.time.LocalTime;
import org.springframework.scheduling.Trigger;
import org.springframework.scheduling.TriggerContext;
public class LastDayOfMonthTrigger implements Trigger {
    private final LocalTime time;
    public LastDayOfMonthTrigger(LocalTime time) {
        this.time = time;
    }
    @Override
    public Date nextExecutionTime(TriggerContext ctx) {
        Date last = ctx.lastScheduledExecutionTime();
        LocalDate date = last == null ? new LocalDate() : new LocalDate(last).plusDays(1);
        LocalDate lastDay = date.dayOfMonth().withMaximumValue();
        return lastDay.toDateTime(time).toDate();
    }
}