how to write Cron expression with initial delay - Quartz in mule?

1.2k Views Asked by At

In my mule flow, currently the cron expression in the Quartz component is 0/10 * * * * ? which triggers for every 10 secs. I need to modify my cron job to work with initial delay of 5 secs? How can i accommodate this by just making change to cron expression?

2

There are 2 best solutions below

0
duddukurimd On BEST ANSWER
<inbound>
            <quartz:inbound-endpoint jobName="myServiceJob" startDelay="5000" repeatInterval="1000">
                <quartz:event-generator-job/>
            </quartz:inbound-endpoint>

2
Victor P. On

The Quartz component has a Start Delay property in milliseconds

<quartz:inbound-endpoint  doc:name="Quartz" jobName="StartFlow" startDelay="5000" repeatInterval="0" cronExpression="0/10 * * * * ?" >