使用 Jackson 将 json 转换为 Java 对象时如何忽略与地图相关的大括号

Question

我正在尝试将 JSON 转换为 Java 对象，但我很难构建 Java 等效对象。

我的JSON看起来像这样

{
    "point1": {
        "x": 1.0,
        "y": 2.0
    },
    "point2": {
        "x": 1.0,
        "y": 2.0
    },
    "point3": {
        "x": 1.0,
        "y": 2.0
    },
    "customobject1": "cust1",
    "customobject2": "cust2"
}

我这里要拿地图取点，因为会有n个点，

public class Test {

    public String getCustomobject1() {
        return customobject1;
    }

    public void setCustomobject1(String customobject1) {
        this.customobject1 = customobject1;
    }

    public String getCustomobject2() {
        return customobject2;
    }

    public void setCustomobject2(String customobject2) {
        this.customobject2 = customobject2;
    }

    Map<String, Point> testing  = new HashMap<>();

    String customobject1;
    String customobject2; 

    public Map<String, Point> getTesting() {
        return testing;
    }

    public void setTesting(Map<String, Point> testing) {
        this.testing = testing;
    }

}

但是我得到了无法识别的属性异常，我知道有一个额外的包装器 ({}) 导致了这个问题，有人可以建议我如何在反序列化时忽略此地图名称 JSON?

注意：我正在工作的实际对象有点复杂，结构相似，我在这里只发布一个原型。

Answer 1

如果您事先不知道密钥，请使用 @JsonAnySetter 映射它们：

Marker annotation that can be used to define a non-static, two-argument method (first argument name of property, second value to set), to be used as a "fallback" handler for all otherwise unrecognized properties found from JSON content. It is similar to XmlAnyElement in behavior; and can only be used to denote a single property per type.

If used, all otherwise unmapped key-value pairs from JSON Object values are added to the property (of type Map or bean).

public class Test  {
  private Map<String, Point> points = new HashMap<>();

  @JsonAnySetter
  public void setPoints(String name, Point value) {
    points.put(name, value);
  }

}

使用 Jackson 将 json 转换为 Java 对象时如何忽略与地图相关的大括号

How to ignore Map related braces while converting json to Java object using Jackson

java

jackson

jackson-databind