At https://github.com/spring-projects/spring-framework/blob/master/spring-context/src/main/kotlin/org/springframework/context/support/BeanDefinitionDsl.kt the comment shows how to define Spring Beans via the new "Functional bean definition Kotlin DSL". I also found https://github.com/sdeleuze/spring-kotlin-functional. However, this example uses just plain Spring and not Spring Boot. Any hint how to use the DSL together with Spring Boot is appreciated.
3 Answers
Spring Boot is based on Java Config, but should allow experimental support of user-defined functional bean declaration DSL via ApplicationContextInitializer support as described here.
In practice, you should be able to declare your beans for example in a Beans.kt file containing a beans() function.
fun beans() = beans {
// Define your bean with Kotlin DSL here
}
Then in order to make it taken in account by Boot when running main() and tests, create an ApplicationContextInitializer class as following:
class BeansInitializer : ApplicationContextInitializer<GenericApplicationContext> {
override fun initialize(context: GenericApplicationContext) =
beans().initialize(context)
}
And ultimately, declare this initializer in your application.properties file:
context.initializer.classes=com.example.BeansInitializer
You will find a full example here and can also follow this issue about dedicated Spring Boot support for functional bean registration.
You can define your beans in *Config.kt file and implement initalize method of ApplicationContextInitializer interface.
override fun initialize(applicationContext: GenericApplicationContext) {
....
}
Some bean definition here.
bean<XServiceImpl>("xService")
bean("beanName") {
BeanConstructor(ref("refBeanName"))
}
@Configurationclass with an@Beanmethod returning the result ofbeans {...}. Then I got the exception"... No qualifying bean of type '...' available ..."when I remove@Serviceand declare the service class inside thebeans {...}lambda above. - Juergen Zimmermann