spring boot cmd line cron expression java.lang.IllegalStateException - spring-boot

Question

I do have an argument with my Spring Boot powered web application. I'm trying to override a #Scheduled cron expression from the command line, but spring responds with a java.lang.IllegalStateException.
Initialization of bean failed;
nested exception is java.lang.IllegalStateException:
Encountered invalid #Scheduled method 'work':
Cron expression must consist of 6 fields (found 1 in "0")
I have a Spring Component with a Scheduled Annotation:
#Scheduled(cron="${myapp.cron}")
public void work() {
...
}
There is an application.properties file like this:
myapp.cron=0 0 1 * * *
I'm bundling the application with maven to a jar file. The application runs on an ubuntu machine with Java 8 as an init task (/etc/init/myapp.conf).
description "My app"
start on runlevel [2345]
stop on runlevel [!2345]
respawn
respawn limit 10 5
setuid <USER>
setgid <GROUP>
script
java -Xms2G -Xmx2G -jar /opt/myapp.jar \
--server.port=4014 \
--server.address=127.0.0.1
--logging.file=/opt/myapp.log \
--logging.level.root=INFO
end script
Until this point everything is fine. Adding the following line gives the above mention exception.
--myapp.cron=0 0 8 * * *
Any ideas? What's wrong?
Cheers,
Kai

Answer 1

Kai,
You're passing in the 0 0 8 * * * as command line parameter in order to get it in the Spring environment. However, your app only gets the first 0 as variable value, hence the exception about the required 6 parts. Surround the value with quotes instead:
--myapp.cron="0 0 8 * * *"

Answer 2

create a xyz.properties file under src/main/resouce folder.
cron.open.status.mgr.schedule=10 15 0 15 * ?
and do something like below
#Component
public class OpenStatusManagerScheduler {
#Scheduled(cron = "${open.status.mgr.schedule}")
public void scheduleStatusTaskWithCronExpression() {}
}