Autowired not working in Scala Spring Boot project - spring

Question

Taking into account the following example where I'm trying to use the Sample configuration bean within SampleStarter to start the service with the bean properly filled. The .scala file has SampleStarter.scala as name, with Sample being defined within that exact same file.
#SpringBootApplication
#Service
object SampleStarter {
#(Autowired #setter)
var sample: Sample = _ // always null, #Autowired not working?
def getInstance() = this
def main(args: Array[String]): Unit = {
SpringApplication.run(classOf[Sample], args: _*)
sample.run()
}
}
#Configuration
#ConfigurationProperties("sample")
#EnableConfigurationProperties
class Sample {
#BeanProperty
var myProperty: String = _
def run(): Unit = { // bean config seems OK, when debugging with #PostConstruct the 'myProperty' value is filled properly
print(myProperty)
}
}
Whenever I hit sample.run() after SpringApplication.run(classOf[Sample], args: _*), sample property is always null. I reckon this has something to do with object in Scala since all their members are static AFAIK. I took this SO question How to use Spring Autowired (or manually wired) in Scala object? as inspiration but still can't make my code to work.
Is there something wrong the way I'm instantiating the classes or is it something related to Scala?
EDIT
Following #Rob's answer, this is what I did trying to replicate his solution.
#SpringBootApplication
#Service
object SampleStarter {
#(Autowired #setter)
var sample: Sample = _
def getInstance() = this
def main(args: Array[String]): Unit = {
SpringApplication.run(classOf[Sample], args: _*)
sample.run()
}
}
#Configuration // declares this class a source of beans
#ConfigurationProperties("sample") // picks up from various config locations
#EnableConfigurationProperties // not certain this is necessary
class AppConfiguration {
#BeanProperty
var myProperty: String = _
#Bean
// Error -> Parameter 0 of constructor in myproject.impl.Sample required a bean of type 'java.lang.String' that could not be found.
def sample: Sample = new Sample(myProperty)
}
class Sample(#BeanProperty var myProperty: String) {
def run(): Unit = {
print(myProperty)
}
}

Answer 1

#Configuration declares a source file that is capable of providing #Beans. This is not what you want. You want/need to have Sample as a POJO class (POSO?) and then have another class "AppConfiguration" (say) annotated with #Configuration that creates an #Bean of type Sample
My scala is very rusty so this may not compile at all but the structure should be roughly correct.
#Configuration // declares this class a source of beans
#ConfigurationProperties("sample") // picks up from various config locations
#EnableConfigurationProperties // not certain this is necessary
class AppConfiguration {
#Bean
def getSample(#BeanProperty myProperty: String): Sample = new Sample(myProperty)
}
class Sample {
var myProperty: String
def Sample(property : String) = {
this.myProperty = myProperty
}
def run(): Unit = {
print(myProperty)
}
}
Now you have a Sample class as an #Bean and it can be #Autowired in where ever necessary.
#SpringBootApplication
#Service
object SampleStarter {
// note its typically better to inject variables into a constructor
// rather than directly into fields as the behaviour is more predictable
#Autowired
var sample: Sample
def getInstance() = this // not required ??
def main(args: Array[String]): Unit = {
SpringApplication.run(classOf[Sample], args: _*)
sample.run()
}
}
FWIW, you can start the SpringBoot application independently and have the logic for #Service entirely separate. An #Service is another type of #Bean and is made available to the rest of the SpringBootApplication.