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Cron expression to trigger on 25 of every month

开发者 https://www.devze.com 2023-03-07 16:59 出处:网络
How to write cron expression to trigger a function on 25th of every month at 9 A.M in the morning? When I execute this code,

How to write cron expression to trigger a function on 25th of every month at 9 A.M in the morning?

When I execute this code,

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,

java.lang.StackOverflowError
    sun.util.calendar.ZoneInfo.getOffsets(Unknown Source)
    sun.util.calendar.ZoneInfo.getOffsets(Unknown Source)
    java.util.GregorianCalendar.compu开发者_高级运维teFields(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)


@Scheduled(cron="0 9 25 1 * ?")

This is on January 1st only, and the time is invalid, you'll want this instead:

@Scheduled(cron="0 0 9 25 * ?")

Reference: CronSequenceGenerator


Corn that will trigger every 25th of each month.

  @Scheduled(cron = "0 0 9 25 * ?")

use this link to validate the expression

https://www.freeformatter.com/cron-expression-generator-quartz.html

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