Question

从教程中我得知这应该有效（简化示例）：

public class Foo {
    private String bar;
    public String getBar() {
        return bar;
    }
    public void setBar(String bar) {
        this.bar = bar;
    }
    public static class Qux {
        private String foobar;
        public String getFoobar() {
            return foobar;
        }
        public void setFoobar(String foobar) {
            this.foobar = foobar;
        }
    }
}
...

String in = "{ \"bar\": \"123\", \"qux\" : {\"foobar\": \"234\"}}";
ObjectMapper mapper = new ObjectMapper();
Foo obj = mapper.readValue(in, Foo.class);

然而，我收到错误

UnrecognizedPropertyException: Unrecognized field "qux" (Class Foo), not marked as ignorable

我正在运行2.2.2

Answer 1

您可以将ObjectMapper配置为忽略您在课堂上找不到的字段

ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

如果没有以这种方式配置，它将在解析时抛出异常，如果它找到了一个它在你指定的类类型上无法识别的字段。

Answer 2

Foo类需要类型为Qux的实例属性才能使自动反序列化工作。当前定义了Foo类的方式，没有目标属性来注入qux JSON对象值。

public class Foo {
   private String bar;

   public String getBar() {
       return bar;
   }

   public void setBar(String bar) {
       this.bar = bar;
   }

   // additional property 
   private Qux qux;

   public Qux getQux() {
       return qux;
   }

   public void setQux(Qux value) {
       qux = value;
   }

   public static class Qux {
       private String foobar;

       public String getFoobar() {
         return foobar;
       }

       public void setFoobar(String foobar) {
           this.foobar = foobar;
       }
    }
}

Answer 3

如果您从Qux

中提取Foo课程，它将会有效

public class Foo {
    private String bar;

    // added this
    private Qux qux;

    public String getBar() {
        return bar;
    }
    public void setBar(String bar) {
        this.bar = bar;
    }

    // added getter and setter
    public Qux getQux() {
        return qux;
    }
    public void setQux(Qux qux) {
        this.qux = bar;
    }
}

public static class Qux {
    private String foobar;
    public String getFoobar() {
        return foobar;
    }
    public void setFoobar(String foobar) {
        this.foobar = foobar;
    }
}

杰克逊反序列化：未被认可的领域

3 个答案: