我有一个Java记录，其中只有一个字段：

public record AggregateId(UUID id) {}

和一个具有AggregateId字段的类（为了可读性，删除了其他字段）

public class Aggregate {
    public final AggregateId aggregateId;
    @JsonCreator
    public Aggregate(
            @JsonProperty("aggregateId") AggregateId aggregateId
    ) {
        this.aggregateId = aggregateId;
    }
}

上面的实现使用给定的示例对JSON进行序列化和反序列化：

ObjectMapper objectMapper = new ObjectMapper();
String content = """
        {
           "aggregateId": {
                "id": "3f61aede-83dd-4049-a6ff-337887b6b807"
            }
        }
        """;
Aggregate aggregate = objectMapper.readValue(content, Aggregate.class);
System.out.println(objectMapper.writeValueAsString(aggregate));

我如何更改Jackson配置以替换JSON：

{
    "aggregateId": "3f61aede-83dd-4049-a6ff-337887b6b807"
}

而不放弃AggregateId的单独类和通过字段访问，不使用getter？
我尝试了@JsonUnwrapper注解，但这导致了抛出

Exception in thread "X" com.fasterxml.jackson.databind.exc.InvalidDefinitionException: 
    Invalid type definition for type `X`: 
        Cannot define Creator parameter as `@JsonUnwrapped`: combination not yet supported at [Source: (String)"{
            "aggregateId": "3f61aede-83dd-4049-a6ff-337887b6b807"
        }"

或

Exception in thread "X" com.fasterxml.jackson.databind.exc.InvalidDefinitionException: 
    Cannot define Creator property "aggregateId" as `@JsonUnwrapped`: 
        combination not yet supported at [Source: (String)"{
            "aggregateId": "3f61aede-83dd-4049-a6ff-337887b6b807"
        }"

• Jackson版本：第2.13.1节 *

dependencies {
    compile "com.fasterxml.jackson.core:jackson-annotations:2.13.1"
    compile "com.fasterxml.jackson.core:jackson-databind:2.13.1"
}

当然，使用自定义的序列化程序/反序列化程序也是可以的，但我正在寻找一种更简单的解决方案，因为我有许多不同的类都有类似的问题。