@Configuration / @ConfigurationProperties doesn't ...

# spring

[Dany]

@Configuration / @ConfigurationProperties doesn't work.

My application.yml

seon:
  uri: 123

My SeonConfiguration class:

@Configuration
@ConfigurationProperties(prefix = "seon")
data class SeonConfiguration(
    var uri: String = ""
)

My main class:

@SpringBootApplication
@ConfigurationPropertiesScan
@EnableConfigurationProperties
class WadzpayServiceApplication
fun main(args: Array<String>) {
    val seonConfiguration: SeonConfiguration = SeonConfiguration()
    print(seonConfiguration.uri) //this line prints null
    runApplication<WadzpayServiceApplication>(*args)
}

I had been looking all the possibile answers on stackoverflow, getting my head crazy but still can't make it work. Any help would be massively appreciated!

[ashmelev]

Well, to start with, Spring has not even had a chance to build it's DAG because you are inspecting the configuration before runApplication(...) is executed. Secondarily, val seonConfiguration: SeonConfiguration = SeonConfiguration() is simply creating a new instance of SeonConfiguration class and not using the one Spring would eventually create - hence the null you are seeing. Lastly, this is not a Kotlin-specific issue you are experiencing. You should spend a few more minutes with the Spring documentation/examples.

[Dany]

@ashmelev, thank you for your insights. I would love to have a foundation understanding of Spring but I was put in a project without any proper preparation and it is quite urgent, therefore I am having trouble to caught up. I am trying to learn about IOC, dependency injection, etc.. as fast as possible, but still having a lot of problems.
Back to the question, after doing a bit of searching I understand your 2 points now. But for the 2nd point, how would I initiate val seonConfiguration to the bean object instead of new object (if I understand it correctly)?

[marzelwidmer]

@Dany what issue U have …. to read the configuration ? try something like this :

@SpringBootApplication
@EnableConfigurationProperties(SeaonConfigurer::class)
class WadzpayServiceApplication

fun main(args: Array<String>) {
    runApplication<WadzpayServiceApplication>(*args)
}


@ConstructorBinding
@ConfigurationProperties(prefix = "seon")
class SeaonConfigurer(var uri: String)

[marzelwidmer]

@Service
class SeonService(private val seaonConfigurer: SeaonConfigurer) {

    fun foo() {
       val bar = seaonConfigurer.host
    }
}

[Dany]

@marzelwidmer, thanks for your example, I tried to follow it, at the end, how would I call the SeonService class? If I create a new object like:
val seonService = SeonService()
then AFAIK, configuration data won't be injected. As far as I understand, I need to call the BEAN object created from Spring, not initiate a new one - I am at this point and stuck

[marzelwidmer]

You have a sample on github ?

[marzelwidmer]

image.png

[marzelwidmer]

or you try to initialise with the DLS? maybe I get it wrong …

[Dany]

Hi, I don't have the sample on GitHub, I am working on a branch on a big project. But here are my code which is following your suggestion.
Here is my application.yml:

seon:
  uri: 123

Here is my SeonConfiguration.kt

@Configuration
@ConfigurationProperties(prefix = "seon")
class SeonConfiguration {
    var uri: String = ""
}

Here is my SeonService.kt

@Service
class SeonService(private val seonConfiguration: SeonConfiguration) {
    fun foo() {
        println(seonConfiguration.uri)
    }
}

Here is my main method - which I don't know how to call the method in SeonService to print out the configuration

@SpringBootApplication
@ConfigurationPropertiesScan
@EnableConfigurationProperties(SeonConfiguration::class)
class WadzpayServiceApplication

fun main(args: Array<String>) {
    runApplication<WadzpayServiceApplication>(*args)
    /*
    What kind of magic here to make the program print 123
    */
}

[marzelwidmer]

Ok my question is why or what you try in the magic part for a use case … ? you try to do something on application start up time ?

[marzelwidmer]

maybe you can do something like

main(args: Array<String>) {
   runApplication<WadzpayServiceApplication>(*args){
      addInitializers(
         beans {
            bean {
               ApplicationRunner { println("ApplicationRunner -----------------> with some magic....") }
            }
         }
      )

   }
}

[Dany]

😂, no, I just wanted it somehow call the method foo from SeonService class, and foo will somehow print the uri from configuration file, which is 123

[marzelwidmer]

image.png

[Dany]

Ok, its probably simple for you, but its still some magic to me, I kinda understand that you call it as a bean object - which was created by Spring, therefore you were able to print the URL, but I failed miserably to achieve that.

Thanks a lot

[marzelwidmer]

Maybe there is also a other way but I think it have to be somehow in a springContext.. I think.

[kqr]

isn't @ConstructorBinding all you were missing? rest is standard dependency injection

[Sourabh Rawat]

The service will be called from the controller. In the controller you have to add service dep.

Although it depends on what you are building? Is a REST API?