如何使用 Jackson 注释将嵌套值映射到属性? [英] How to map a nested value to a property using Jackson annotations?

问题描述

假设我正在调用一个 API，该 API 以产品的以下 JSON 进行响应:

Let's say I'm making a call to an API that responds with the following JSON for a product:

{
  "id": 123,
  "name": "The Best Product",
  "brand": {
     "id": 234,
     "name": "ACME Products"
  }
}

我可以使用 Jackson 注释很好地映射产品 ID 和名称:

I'm able to map the product id and name just fine using Jackson annotations:

public class ProductTest {
    private int productId;
    private String productName, brandName;

    @JsonProperty("id")
    public int getProductId() {
        return productId;
    }

    public void setProductId(int productId) {
        this.productId = productId;
    }

    @JsonProperty("name")
    public String getProductName() {
        return productName;
    }

    public void setProductName(String productName) {
        this.productName = productName;
    }

    public String getBrandName() {
        return brandName;
    }

    public void setBrandName(String brandName) {
        this.brandName = brandName;
    }
}

然后使用 fromJson 方法创建产品:

And then using the fromJson method to create the product:

  JsonNode apiResponse = api.getResponse();
  Product product = Json.fromJson(apiResponse, Product.class);

但现在我想弄清楚如何获取品牌名称，这是一个嵌套属性.我希望这样的事情会奏效:

But now I'm trying to figure out how to grab the brand name, which is a nested property. I was hoping that something like this would work:

    @JsonProperty("brand.name")
    public String getBrandName() {
        return brandName;
    }

但当然没有.有没有一种简单的方法可以使用注释来完成我想要的操作?

But of course it didn't. Is there an easy way to accomplish what I want using annotations?

我试图解析的实际 JSON 响应非常复杂，我不想为每个子节点创建一个全新的类，即使我只需要一个字段.

The actual JSON response I'm trying to parse is very complex, and I don't want to have to create an entire new class for every sub-node, even though I only need a single field.

推荐答案

你可以这样实现:

String brandName;

@JsonProperty("brand")
private void unpackNameFromNestedObject(Map<String, String> brand) {
    brandName = brand.get("name");
}

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