Spring @Scheduled annotation random delay

Question

I am using the @Scheduled annotation from Spring framework to invoke a method. But I have multiple nodes in my setup and I do not want them to all run at exactly the same time. So I'd like to set a random value to the initial delay to offset them from each other.

import org.springframework.scheduling.annotation.Scheduled;

@Scheduled(fixedRate = 600000, initialDelay = <random number between 0 and 10 minutes> )

Unfortunately, I am only allowed to use a constant expression here. Is there any other way around this? I thought of using Spring expression language.

Related: [Spring @Scheduled to be started every day at a random minute between 4:00AM and 4:30AM](https://stackoverflow.com/questions/41016433/spring-scheduled-to-be-started-every-day-at-a-random-minute-between-400am-and) — sleske, May 12 '23 at 09:58

score 20 · Accepted Answer · edited Aug 23 '18 at 07:25

20

You can configure the initialDelay through Spring Expression Language:

@Scheduled(fixedRate = 600000, initialDelayString = "#{ T(java.util.concurrent.ThreadLocalRandom).current().nextInt(10*60*1000) }" )

I don't have an IDE to test that code right now, so you may need to adapt that a bit.

edited Aug 23 '18 at 07:25

Eduard Wirch

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answered Jan 10 '15 at 01:18

sk_

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Don't SpEl commands need to be in strings? I've only seen them used in string based values. This unfortunately is expecting a long value. I tried it unquoted and got unresolved compilation problems. I tried it quoted as a string and the scheduler never got invoked but no error messages. – troymass Jan 10 '15 at 05:51
Have you noticed the use of `initialDelayString` instead of `initialDelay`? The latter takes a long, the first a string which can be an expression. – M. Deinum Jan 10 '15 at 07:34
The example does not quite work. First, it must quoted or the compiler will complain. But even then, all I get is null. Is there som kind of configuration that I need to do? – girgen Jun 08 '15 at 15:26
Confirmed with multiple Spring versions that this does not work – anupash Aug 01 '18 at 21:19

score 18 · Answer 2 · answered Jul 12 '16 at 02:01

18

To make the initial delay randomly somewhere between 0 and the fixedRate try this:

@Scheduled(fixedDelayString = "${some.delay}", initialDelayString = "${random.int(${some.delay})}")

Where you define some.delay (but pick a more suitable name) as 10 minutes as a property like so in your application.properties or equivalent.

some.delay = 600000

Of course if you want to be lazy and hard code it you can always just use ${random.int(600000)}

answered Jul 12 '16 at 02:01

mekazu

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Just wanted to add this alternative answer as the 'correct' answer isn't really helpful and yet this page is high up in search results. – mekazu Jul 12 '16 at 02:03
1

Confirmed with multiple Spring versions that this does not work – anupash Aug 01 '18 at 21:19
I have used this exact solution in multiple projects, and it worked great. Have not tried it with most recent versions of spring. – Samuel Lindblom Jul 28 '21 at 15:05

yglodt · Answer 3 · 2017-04-02T17:39:42.573

5

In this working example, the random delay will be between 5 and 10 seconds.

@Scheduled(fixedDelayString = "#{new Double((T(java.lang.Math).random() + 1) * 5000).intValue()}")

edited Apr 02 '17 at 17:39

answered Jan 23 '17 at 21:26

yglodt

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score 2 · Answer 4 · edited Dec 13 '21 at 05:56

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Keep in mind, that the initialDelayString is evaluated only once at startup and then this same values is used whenever the job is scheduled.

See org.springframework.scheduling.annotation.ScheduledAnnotationBeanPostProcessor#processScheduled

edited Dec 13 '21 at 05:56

Yu Hao

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answered Jul 25 '16 at 07:02

Stefan Isele - prefabware.com

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Do you know a way to have the scheduler not only have a constant random value evaluated at startup, but a random value between each triggering? – yglodt Jan 23 '17 at 21:36

slorinc · Answer 5 · 2020-11-07T11:59:09.913

This should work

@Scheduled(fixedRate = 600000, initialDelay = "#{new java.util.Random().nextInt(700)}")

Verified it with this simple test:

    @Test
    public void testSpEL() {
       ExpressionParser parser = new SpelExpressionParser();
       Expression exp = parser.parseExpression("new java.util.Random().nextInt(500)");
       Integer value =(Integer) exp.getValue();
       Assertions.assertThat(value).isNotNull();
}

score 0 · Answer 6 · answered Aug 28 '19 at 11:40

0

In kotlin this works:

@Component
class MyJob {
   companion object {
      const val INTERVAL = 24*3600*1000L // once a day
   }

   @Scheduled(fixedRate = INTERVAL, initialDelayString = "\${random.long($INTERVAL)}")
   fun doDaily() {
      ...
   }
}

answered Aug 28 '19 at 11:40

Milo van der Zee

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score 0 · Answer 7 · answered Oct 19 '22 at 11:51

0

The shortest way I found is:

@Scheduled(initialDelayString = "#{T(Math).round(T(Math).random()*600000)}")

answered Oct 19 '22 at 11:51

Mert Ülkgün

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score -6 · Answer 8 · answered May 19 '17 at 09:59

-6

Or you could just add Thread.sleep(...) at the end of your function.

@Scheduled(fixedDelay = 10000)
public void replicateData() {

    ... do stuff ...

    try {
        Thread.sleep(RandomUtils.nextLong(1000, 10000));
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
}

answered May 19 '17 at 09:59

Vincnetas

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It would need to be a random delay at the start of the function. And there would need to be code to make that delay only the first time. That might work but probably a bit clunky – troymass May 19 '17 at 17:11

Spring @Scheduled annotation random delay

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