How to use “Functional bean definition Kotlin DSL”

问题:

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.

回答1:

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.

回答2:

Another way to do it in Spring Boot would be :

 fun main(args: Array<String>){

 SpringApplicationBuilder().initializers( beans {

    // Define your bean with Kotlin DSL here


 }).sources(MainClass::class.java).run(*args);