Controller not executed in my Spring Boot application - spring

Question

I tried to create simple Hello World app in Spring Boot using kotlin but IntelliJ IDE shows me warning that my controller class is never used and the specified endpoint does not work as well. I can't figure out what to do with it.
I created the app with the Boot Initializr and the structure looks like this:
kotlin/
com.myapp.school/
Application.kt
controller/
HelloController.kt
resources/
static/
templates/
hello.html
And here is the code for Application.kt:
package com.myapp.school
import org.springframework.boot.autoconfigure.SpringBootApplication
import org.springframework.boot.runApplication
#SpringBootApplication
class Application
fun main(args: Array<String>) {
runApplication<Application>(*args)
}
Last, I have this simple controller with one method:
package com.myapp.school.controller
import org.springframework.stereotype.Controller
import org.springframework.ui.Model
import org.springframework.web.bind.annotation.GetMapping
#Controller
class HelloController
#GetMapping("/hello")
fun hello(): String {
System.out.println("Hello from controller")
return "hello"
}
Going to localhost:8080/hello shows whitelabel error page with 404 status. I read that Spring prints registered endpoints into the console at startup but I haven't found such message.
Can somebody tell me what is wrong?
Thank you

Answer 1

I think your problem is that you have a top level class with no body (HelloController) and a top level function (hello). You have to put curly braces to make sure hello is a member of HelloController.
You have this:
#Controller
class HelloController
#GetMapping("/hello")
fun hello(): String {
System.out.println("Hello from controller")
return "hello"
}
It needs to be like this, so hello belongs in HelloController, not at the same level:
#Controller
class HelloController {
#GetMapping("/hello")
fun hello(): String {
System.out.println("Hello from controller")
return "hello"
}
}
Also, change System.out.println to just println to be more Kotlin-like.