java 每月 25 日触发的 Cron 表达式
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时间：2020-10-30 14:14:44 来源:igfitidea点击:
Cron expression to trigger on 25 of every month
提问by elle
How to write cron expression to trigger a function on 25th of every month at 9 A.M in the morning?
如何编写cron表达式在每个月的25日早上9点触发一个函数？
When I execute this code,
当我执行这段代码时，
import org.springframework.scheduling.annotation.Scheduled;
import org.springframework.stereotype.Service;
@Service
public class PayrollSchedulerImpl implements PayrollScheduler{
@Scheduled(cron="0 9 25 1 * ?")
public void calculateSalaryScheduled()
{
calculateSalary();
}
public void calculateSalary()
{
/* */
}
}
I get the error,
我得到错误，
java.lang.StackOverflowError
sun.util.calendar.ZoneInfo.getOffsets(Unknown Source)
sun.util.calendar.ZoneInfo.getOffsets(Unknown Source)
java.util.GregorianCalendar.computeFields(Unknown Source)
java.util.GregorianCalendar.computeTime(Unknown Source)
java.util.Calendar.updateTime(Unknown Source)
java.util.Calendar.complete(Unknown Source)
java.util.Calendar.get(Unknown Source)
org.springframework.scheduling.support.CronSequenceGenerator.doNext(CronSequenceGenerator.java:130)
回答by Sean Patrick Floyd
@Scheduled(cron="0 9 25 1 * ?")
This is on January 1st only, and the time is invalid, you'll want this instead:
这只是在 1 月 1 日，时间无效，你会想要这个：
@Scheduled(cron="0 0 9 25 * ?")
Reference:CronSequenceGenerator
