'一对多' / '多对多' 属性值类型不应该是'?延伸'答案

【问题标题】：'one to many' / 'many to many' attribute value type should not be '? extends''一对多' / '多对多' 属性值类型不应该是'?延伸'
【发布时间】：2020-07-03 00:40:29
【问题描述】：

我正在尝试使用 Kotlin 中的注释定义一个简单的 Hibernate 映射。但是我的多对多关系没有按预期工作。在 IntelliJ IDEA 中导致以下错误：

'one to many' / 'many to many' 属性值类型不应该是'?扩展 PersonAddress/Person'

我的代码库：

@Entity
open class Person(
        @Id
        @GeneratedValue(strategy = GenerationType.SEQUENCE)
        var id: Long = 0L,

        @OneToMany(mappedBy = "person")
        var addresses: Set<PersonAddress> = setOf() //Fail

)

@Entity
open class Address(
        @Id
        @GeneratedValue(strategy = GenerationType.SEQUENCE)
        var id: Long = 0L,

        @OneToMany(mappedBy = "address")
        var persons: Set<PersonAddress> = setOf() //Fail

)

@Entity
open class PersonAddress(
        @Id
        var id: Long,

        @ManyToOne
        @JoinColumn(name = "person_id")
        var person: Person,

        @ManyToOne
        @JoinColumn(name = "address_id")
        var address: Address
) : Serializable

所以我认为这可能是由 Join-Table 引起的错误，所以我对 ManyToMany Relation 进行了同样的尝试：

 @ManyToMany(mappedBy = "addresses")
var persons: Set<Person> = setOf()

但同样的错误发生在这里。

Application.kt

@SpringBootApplication
class Application {
    fun main(args: Array<String>) {
        runApplication<Application>(*args)
    }
}

build.gradle.kts

plugins {
    java
    id("org.springframework.boot") version "2.2.5.RELEASE"
    id("io.spring.dependency-management") version "1.0.9.RELEASE"
    kotlin("jvm")
    kotlin("plugin.spring") version "1.3.70"
    kotlin("plugin.jpa")
    kotlin("plugin.allopen") version "1.3.70"
}

val developmentOnly by configurations.creating
configurations {
    runtimeClasspath {
        extendsFrom(developmentOnly)
    }
}

repositories {
    mavenCentral()
}

dependencies {
    implementation("org.springframework.boot:spring-boot-starter-data-jpa")
    implementation("org.springframework.boot:spring-boot-starter-validation")
    implementation("org.jetbrains.kotlin:kotlin-reflect")
    implementation("org.jetbrains.kotlin:kotlin-stdlib-jdk8")
    runtimeOnly("com.h2database:h2")
    developmentOnly("org.springframework.boot:spring-boot-devtools")
    testImplementation("org.springframework.boot:spring-boot-starter-test") {
        exclude(group = "org.junit.vintage", module = "junit-vintage-engine")
    }
}

allOpen {
    annotation("javax.persistence.Entity")
    annotation("javax.persistence.Embeddable")
    annotation("javax.persistence.MappedSuperclass")
}

configure<JavaPluginConvention> {
    sourceCompatibility = JavaVersion.VERSION_1_8
}
tasks {
    compileKotlin {
        kotlinOptions.jvmTarget = "1.8"
    }
    compileTestKotlin {
        kotlinOptions.jvmTarget = "1.8"
    }
}

tasks.withType<org.jetbrains.kotlin.gradle.tasks.KotlinCompile> {
    kotlinOptions {
        freeCompilerArgs = listOf("-Xjsr305=strict")
        jvmTarget = "1.8"
    }
}

tasks.withType<Test> {
    useJUnitPlatform()
}

编辑 1：我通过使用摆脱了错误：

var persons: MutableList<PersonAddress> = mutableListOf()

但是浏览我的代码时，错误发生在昨天运行良好的属性上。

【问题讨论】：

标签： spring hibernate spring-boot jpa kotlin

【解决方案1】：

所以回答我自己的问题：

由于我正在使用open 类，因此 List 必须是一个可变集合。错误信息可能会更好。

@OneToMany(mappedBy = "address")
var persons: MutableList<PersonAddress> = mutableListOf()

【讨论】：