@jsonanysetter,@jsonanygetter with DSL Json(De)/序列化不取值（始终为空）

Question

  @CompiledJson
  public class Name {

String name;
public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

Map<String,String> properties = new HashMap<String,String>();


public Name() {
    // TODO Auto-generated constructor stub
}

@JsonAnyGetter
public Map<String, String> get() {
    return this.properties;
}

@JsonAnySetter
public void set(String key, String value) {
    this.properties.put(key, value);
}

使用DSLJson serialize（）和deserialize（）方法解/序列化。我也没有看到任何错误，但在JSON中properties仍然为null。我怀疑DSL JSON是否支持Jackson注释。：/

带有DSL、Json和Jackson注释的Spring Boot应用程序

更新我想解析MyClass，它是RootClass的一部分：

 @Compiledjson
 public class RootClass {

private String id;
private List<MyClass> myclass;
private AnotherCLass class2;

//getters and setter here
}

 @CompiledJson
 public class MyClass implements JsonObject {

 private String name;

 private Map<String, String> properties; //want this to behave like Jackson's       @JsonAnySetter/Getter annotation.
 //The implementation of MapConverter serializer you mentioned below.
}

{
"id" : "1234",
"myclass" :
[
{
"name" : "abcd",
//any dynamic properties I want to add will go here
 "test" : "test1",
 "anything" : "anything"
}
],
"class2" : "test5"
}

谢谢：）

Answer 1

DSL-JSON不支持这样的get（）/set（string,string）方法对。它确实理解map ，所以如果您公开 properties，它就会在这方面起作用。但不是在这种情况下。

从V1.1开始，您有两个解决此类问题的选项，这两个选项都包含在示例项目中

如果您希望重用现有转换器，您的解决方案可以如下所示：

public static class MyClass {

    private String name;
    private Map<String, String> properties;

    @JsonConverter(target = MyClass.class)
    public static class MyClassConverter {
        public static final JsonReader.ReadObject<MyClass> JSON_READER = new JsonReader.ReadObject<MyClass>() {
            public MyClass read(JsonReader reader) throws IOException {
                Map<String, String> properties = MapConverter.deserialize(reader);
                MyClass result = new MyClass();
                result.name = properties.get("name");
                result.properties = properties;
                return result;
            }
        };
        public static final JsonWriter.WriteObject<MyClass> JSON_WRITER = new JsonWriter.WriteObject<MyClass>() {
            public void write(JsonWriter writer, MyClass value) {
                MapConverter.serialize(value.properties, writer);
            }
        };
    }
}