你可以写一个序列化类来实现:

public class Symbol

{
     private String symbol;

     private String name;

     public String getSymbol() {
        return symbol;
    }
    public void setSymbol(String symbol) {
        this.symbol = symbol;
    }    
    public String getName() {
        return name;
    }    
    public void setName(String name) {
        this.name = name;
    }
}

public class SymbolJsonSerializer extends JsonSerializer<Symbol> {

    @Override
    public void serialize(Symbol symbol, JsonGenerator jgen, SerializerProvider serializers) throws IOException, JsonProcessingException {
        jgen.writeStartObject();

        jgen.writeStringField("symbol", symbol.getSymbol());
         //Changed name to full_name as the field name of Json string
        jgen.writeStringField("full_name", symbol.getName());
        jgen.writeEndObject(); 
    }
}

            ObjectMapper mapper = new ObjectMapper();

            SimpleModule module = new SimpleModule();
            module.addSerializer(Symbol.class, new SymbolJsonSerializer());
            mapper.registerModule(module); 

            //only convert non-null field, option...
            mapper.setSerializationInclusion(Include.NON_NULL); 

            String jsonString = mapper.writeValueAsString(symbolList);

我将两个不同的getter /setter对绑定到一个变量:

class Coordinates{
    int red;

    @JsonProperty("red")
    public byte getRed() {
      return red;
    }

    public void setRed(byte red) {
      this.red = red;
    }

    @JsonProperty("r")
    public byte getR() {
      return red;
    }

    public void setR(byte red) {
      this.red = red;
    }
}

刚刚测试，这是有效的:

public class Coordinates {
    byte red;

    @JsonProperty("r")
    public byte getR() {
      return red;
    }

    @JsonProperty("red")
    public void setRed(byte red) {
      this.red = red;
    }
}

其思想是方法名应该是不同的，因此jackson将其解析为不同的字段，而不是一个字段。

下面是测试代码:

Coordinates c = new Coordinates();
c.setRed((byte) 5);

ObjectMapper mapper = new ObjectMapper();
System.out.println("Serialization: " + mapper.writeValueAsString(c));

Coordinates r = mapper.readValue("{\"red\":25}",Coordinates.class);
System.out.println("Deserialization: " + r.getR());

结果:

Serialization: {"r":5}
Deserialization: 25

他们必须将此作为一个特性包含进来，因为现在为getter和setter设置不同的@JsonProperty会得到您所期望的结果(在同一字段的序列化和反序列化期间，属性名不同)。Jackson 2.6.7版本

可以有普通的getter/setter对。您只需要在@JsonProperty中指定访问模式

下面是单元测试:

public class JsonPropertyTest {

  private static class TestJackson {

    private String color;

    @JsonProperty(value = "device_color", access = JsonProperty.Access.READ_ONLY)
    public String getColor() {
      return color;
    };

    @JsonProperty(value = "color", access = JsonProperty.Access.WRITE_ONLY)
    public void setColor(String color) {
      this.color = color;
    }

  }

  @Test
  public void shouldParseWithAccessModeSpecified() throws Exception {
    String colorJson = "{\"color\":\"red\"}";
    ObjectMapper mapper = new ObjectMapper();
    TestJackson colotObject = mapper.readValue(colorJson, TestJackson.class);

    String ser = mapper.writeValueAsString(colotObject);
    System.out.println("Serialized colotObject: " + ser);
  }
}

我得到的输出如下:

Serialized colotObject: {"device_color":"red"}

使用Jackson 2.9+引入的@JsonAlias进行注释，而不用在要用多个别名(json属性的不同名称)反序列化的项上提到@JsonProperty就可以了。

我使用com.fasterxml.jackson.annotation.JsonAlias与com.fasterxml.jackson.databind.ObjectMapper作为我的用例来实现包的一致性。

例如:

@Data
@Builder
public class Chair {

    @JsonAlias({"woodenChair", "steelChair"})
    private String entityType;

}


@Test
public void test1() {

   String str1 = "{\"woodenChair\":\"chair made of wood\"}";
   System.out.println( mapper.readValue(str1, Chair.class));
   String str2 = "{\"steelChair\":\"chair made of steel\"}";
   System.out.println( mapper.readValue(str2, Chair.class));

}

工作得很好。