spring @scheduled cron with variable

Question

in our system we use a settings class to point to the properties file, depends on which eve it loads different property file. To access specific property, we call 'Settings.getString('property_name_here')'.

In my code, i loaded the @scheduled cron expression to a variable and try to passed to the @scheduled annotation, but it won't work,

here is my code: in properties file:

cron.second=0
cron.min=1
cron.hour=14

in constructor i have:

this.cronExpression = new StringBuilder()
            .append(settings.getString("cron.second"))
            .append(" ")
            .append(settings.getString("cron.min"))
            .append(" ")
            .append(settings.getString("cron.hour"))
            .append(" ")
            .append("*").append(" ").append("*").append(" ").append("*")
            .toString();

which create a String of "0 1 14 * * *", it's a valid con expression

in the scheduled task i have:

@Scheduled(cron = "${this.cronExpression}")
public void scheduleTask() throws Exception {
         ....
 }

when I ran the code it complaint: Caused by: java.lang.IllegalStateException: Encountered invalid @Scheduled method 'scheduleTask': Cron expression must consist of 6 fields (found 1 in "${this.cronExpression}")

then I changed this.cronExpression to a list of String:

    this.cronExpression = Lists.newArrayList();
    this.cronExpression.add(settings.getString("cron.second"));
    this.cronExpression.add(settings.getString("cron.min"));
    this.cronExpression.add(settings.getString("cron.hour"));
    this.cronExpression.add("*");
    this.cronExpression.add("*");
    this.cronExpression.add("*");

but still got the same error, so what exactly is the cron expression supposed to be?

Answer 1

What I did and it worked for me:

In properties file:

my.cron.expression=0 3 * * * ?

In code:

@Scheduled(cron = "${my.cron.expression}")
public void scheduleTask(){
  ...
}