Kotlin JPA实体ID [英] Kotlin JPA entity ID

问题描述

定义JPA实体ID的"kotlin方法"是什么?

What is "the kotlin way" to define JPA entity ID?

@Entity
data class User (
        @Id @GeneratedValue
        var id: Long? = null,
        ...
)

还是有更好的方法来避免可为空的id?

Or is there any better one to avoid nullable id?

推荐答案

您可以使用0值而不是空值.

You can use a 0 value rather than a null value.

@Entity
data class User (
        @Id @GeneratedValue
        var id: Long = 0,
        ...
)

自动生成仍应找到下一个序列.

Autogeneration should still find the next sequence.

Kotlin编译为Java，Java具有原始类型long和类Long

Kotlin compiles to Java, which has both, a primitive type long and a Class Long

按照 Java 11.1.21 Id注释部分中的持久性规范 都可以用于Id:

As per the Java Persistence Specification in section 11.1.21 Id Annotation both can be used for the Id:

应用了Id注释的字段或属性应该是一个 以下类型之一:任何 Java原始类型；任何原始包装器类型； java.lang.String; java.util.Date; java.sql.Date; java.math.BigDecimal; java.math.BigInteger [109].

The field or property to which the Id annotation is applied should be one of the following types: any Java primitive type; any primitive wrapper type; java.lang.String; java.util.Date; java.sql.Date; java.math.BigDecimal; java.math.BigInteger[109].

使用Class优于原始类型，因为null具有更明确的含义.但是从规范来看，两者都是可能的，并且您必须确定天气情况，而不是jpa风格还是相反，您更倾向于Kotlins nullsafety.

There is an advantage in using the Class over the primitive, as null has a more unambiguous meaning. But from the spec both are possible and you have to decide weather you favor Kotlins nullsafety over the the jpa style or the other way around.

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