3

I am using the package "cron": "^1.7.1".

I want to finish a task that can take longer than the scheduled cron-job.

Find below my minimum viable example:

const CronJob = require('cron').CronJob;

console.log('Before job instantiation');
const job3 = new CronJob(
  '*/2 * * * * *', async () => {
    if (job3.taskRunning) {
      return
    }

    try {
      //run longer task here
      await setTimeout(() => {
        const d = new Date();
        console.log('JOB 3 - ', d);
        job3.taskRunning = true
      }, 6000);
    } catch (err) {
      console.log(err);
    }
    job3.taskRunning = false
  }
)
console.log('After job instantiation');
job3.start();

As you can see my job runs every 2 seconds and prints:

JOB 3 -  2019-09-01T17:06:22.006Z
JOB 3 -  2019-09-01T17:06:24.001Z
JOB 3 -  2019-09-01T17:06:26.002Z
JOB 3 -  2019-09-01T17:06:28.001Z

However, I would like to get the message only every 6 seconds as the task needs to run 6 seconds.

Any suggestions what I am doing wrong?

Carol.Kar
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1 Answers1

4

Correctly initialising run identifier fixes the problem. Create global variable taskRunning initialised to saying not running. Then just before calling long running task make it in run state. Before long running task finishes again set it to not running. Added extra console log in case next fire time occurs before long running task is completed and returning without executing it.

const CronJob = require('cron').CronJob;
taskRunning=false
console.log('Before job instantiation');
const job3 = new CronJob(
  '*/2 * * * * *', async () => {
    if (taskRunning) {
      console.log('returning')
      return
    }
    taskRunning=true
    try {
      //run longer task here
      await setTimeout(() => {
        const d = new Date();
        console.log('JOB 3 - ', d);
        taskRunning = false
      }, 6000);
    } catch (err) {
      console.log(err);
    }
    
  }
)
console.log('After job instantiation');
job3.start();
user1737906
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