[FIXED] Hot to initialize MongoRepository to load random data in SprintBoot using Kotlin

Issue

I have a simple test case (because I didn't find another way to do it) to load my database with some random data:

package com.springbootapirest.app

import com.springbootapirest.app.models.Dog
import com.springbootapirest.app.repositories.DogRepository
import org.jeasy.random.EasyRandom
import org.junit.jupiter.api.Test
import org.springframework.beans.factory.annotation.Autowired
import java.util.stream.Collectors


internal class InitializationTest() {

    @Autowired
    private lateinit var dogRepository: DogRepository

    @Test
    fun initializeDatabase() {
        val generator = EasyRandom()
        val dogs: List<Dog> = generator.objects(Dog::class.java, 3)
            .collect(Collectors.toList())
        println(dogs);
       dogs.forEach { dog ->
           this.dogRepository.save(dog);
       }
    }
}

But when I try to execute with ./gradlew test --tests InitializationTest

The error fragment is kotlin.UninitializedPropertyAccessException: lateinit property dogRepository has not been initialized

And I don't find how I must initialize the MongoRepository to work exactly the same way as if the Spring Boot API service was running.

Any other simple approach or help will be welcome, thanks.

Solution

At the end was to easy as add @SpringBootTest in the header of the class:

@SpringBootTest
internal class InitializationTest() {

    @Autowired
    private lateinit var dogRepository: DogRepository

    @Test
    fun initializeDatabase() {
        val generator = EasyRandom()
        val dogs: List<Dog> = generator.objects(Dog::class.java, 3)
            .collect(Collectors.toList())
        println(dogs);
       dogs.forEach { dog ->
           this.dogRepository.save(dog);
       }
    }
}

Answered By - Juan Antonio
Answer Checked By - Candace Johnson (JavaFixing Volunteer)