108

有没有办法使用杰克逊 JSON 处理器进行自定义字段级序列化？例如，我想上课

public class Person {
    public String name;
    public int age;
    public int favoriteNumber;
}

序列化为以下 JSON：

{ "name": "Joe", "age": 25, "favoriteNumber": "123" }

请注意，age=25 被编码为数字，而 favoriteNumber=123 被编码为字符串。开箱即用的杰克逊编组int到一个数字。在这种情况下，我希望将 favoriteNumber 编码为字符串。

4

7 回答 7

126

您可以按如下方式实现自定义序列化程序：

public class Person {
    public String name;
    public int age;
    @JsonSerialize(using = IntToStringSerializer.class, as=String.class)
    public int favoriteNumber:
}


public class IntToStringSerializer extends JsonSerializer<Integer> {

    @Override
    public void serialize(Integer tmpInt, 
                          JsonGenerator jsonGenerator, 
                          SerializerProvider serializerProvider) 
                          throws IOException, JsonProcessingException {
        jsonGenerator.writeObject(tmpInt.toString());
    }
}

Java 应该为您处理从int到的自动装箱Integer。

于 2012-08-21T00:25:17.300 回答
61

Jackson-databind (至少 2.1.3) 提供了特殊的ToStringSerializer( com.fasterxml.jackson.databind.ser.std.ToStringSerializer)

例子：

public class Person {
    public String name;
    public int age;
    @JsonSerialize(using = ToStringSerializer.class)
    public int favoriteNumber:
}
于 2013-02-14T11:24:56.757 回答
15

为该字段添加一个带@JsonProperty注释的 getter，它返回 a ：StringfavoriteNumber

public class Person {
    public String name;
    public int age;
    private int favoriteNumber;

    public Person(String name, int age, int favoriteNumber) {
        this.name = name;
        this.age = age;
        this.favoriteNumber = favoriteNumber;
    }

    @JsonProperty
    public String getFavoriteNumber() {
        return String.valueOf(favoriteNumber);
    }

    public static void main(String... args) throws Exception {
        Person p = new Person("Joe", 25, 123);
        ObjectMapper mapper = new ObjectMapper();
        System.out.println(mapper.writeValueAsString(p)); 
        // {"name":"Joe","age":25,"favoriteNumber":"123"}
    }
}
于 2012-08-21T00:13:58.287 回答
15

jackson-annotations 提供@JsonFormat了无需编写自定义序列化程序即可处理大量自定义的功能。

例如，STRING为数字类型的字段请求形状会将数值输出为字符串

public class Person {
    public String name;
    public int age;
    @JsonFormat(shape = JsonFormat.Shape.STRING)
    public int favoriteNumber;
}

将产生所需的输出

{"name":"Joe","age":25,"favoriteNumber":"123"}
于 2017-08-15T13:51:16.760 回答
11

如果你不想用注释污染你的模型并且想要执行一些自定义操作，你可以使用 mixins。

ObjectMapper mapper = new ObjectMapper();
SimpleModule simpleModule = new SimpleModule();
simpleModule.setMixInAnnotation(Person.class, PersonMixin.class);
mapper.registerModule(simpleModule);

覆盖年龄：

public abstract class PersonMixin {
    @JsonSerialize(using = PersonAgeSerializer.class)
    public String age;
}

随年龄做任何你想做的事：

public class PersonAgeSerializer extends JsonSerializer<Integer> {
    @Override
    public void serialize(Integer integer, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException {
        jsonGenerator.writeString(String.valueOf(integer * 52) + " months");
    }
}
于 2019-04-05T09:34:49.013 回答
3

在@JsonView的帮助下，我们可以决定要序列化的模型类的字段满足最低标准（我们必须定义标准），就像我们可以拥有一个具有 10 个属性的核心类，但只能序列化 5 个客户需要的属性只要

通过简单地创建以下类来定义我们的视图：

public class Views
{
    static class Android{};
    static class IOS{};
    static class Web{};
}

带视图的注释模型类：

public class Demo 
{
    public Demo() 
    {
    }

@JsonView(Views.IOS.class)
private String iosField;

@JsonView(Views.Android.class)
private String androidField;

@JsonView(Views.Web.class)
private String webField;

 // getters/setters
...
..
}

现在我们必须通过简单地从 spring 扩展 HttpMessageConverter 类来编写自定义 json 转换器：

    public class CustomJacksonConverter implements HttpMessageConverter<Object> 
    {
    public CustomJacksonConverter() 
        {
            super();
        //this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView(Views.ClientView.class));
        this.delegate.getObjectMapper().configure(MapperFeature.DEFAULT_VIEW_INCLUSION, true);
        this.delegate.getObjectMapper().setSerializationInclusion(Include.NON_NULL);

    }

    // a real message converter that will respond to methods and do the actual work
    private MappingJackson2HttpMessageConverter delegate = new MappingJackson2HttpMessageConverter();

    @Override
    public boolean canRead(Class<?> clazz, MediaType mediaType) {
        return delegate.canRead(clazz, mediaType);
    }

    @Override
    public boolean canWrite(Class<?> clazz, MediaType mediaType) {
        return delegate.canWrite(clazz, mediaType);
    }

    @Override
    public List<MediaType> getSupportedMediaTypes() {
        return delegate.getSupportedMediaTypes();
    }

    @Override
    public Object read(Class<? extends Object> clazz,
            HttpInputMessage inputMessage) throws IOException,
            HttpMessageNotReadableException {
        return delegate.read(clazz, inputMessage);
    }

    @Override
    public void write(Object obj, MediaType contentType, HttpOutputMessage outputMessage) throws IOException, HttpMessageNotWritableException 
    {
        synchronized(this) 
        {
            String userAgent = ((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest().getHeader("userAgent");
            if ( userAgent != null ) 
            {
                switch (userAgent) 
                {
                case "IOS" :
                    this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView(Views.IOS.class));
                    break;
                case "Android" :
                    this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView(Views.Android.class));
                    break;
                case "Web" :
                    this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView( Views.Web.class));
                    break;
                default:
                    this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView( null ));
                    break;
                }
            }
            else
            {
                // reset to default view
                this.delegate.getObjectMapper().setConfig(this.delegate.getObjectMapper().getSerializationConfig().withView( null ));
            }
            delegate.write(obj, contentType, outputMessage);
        }
    }

}

现在需要告诉spring使用这个自定义的json转换，只需将它放在dispatcher-servlet.xml中

<mvc:annotation-driven>
        <mvc:message-converters register-defaults="true">
            <bean id="jsonConverter" class="com.mactores.org.CustomJacksonConverter" >
            </bean>
        </mvc:message-converters>
    </mvc:annotation-driven>

这就是您能够决定要序列化哪些字段的方式。

于 2016-07-12T06:25:04.937 回答
1

您可以在 mixin 中内联创建自定义序列化程序。然后用它注释一个字段。请参阅下面的示例，该示例将“ - something else ”附加到 lang 字段。这有点骇人听闻- 如果您的序列化程序需要诸如存储库之类的东西或 spring 注入的任何东西，这将是一个问题。可能最好使用自定义反序列化器/序列化器而不是 mixin。

package com.test;

import com.fasterxml.jackson.annotation.JsonAutoDetect;
import com.fasterxml.jackson.annotation.JsonAutoDetect.Visibility;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonPropertyOrder;
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import com.test.Argument;
import java.io.IOException;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
//Serialize only fields explicitly mentioned by this mixin.
@JsonAutoDetect(
    fieldVisibility = Visibility.NONE,
    setterVisibility = Visibility.NONE,
    getterVisibility = Visibility.NONE,
    isGetterVisibility = Visibility.NONE,
    creatorVisibility = Visibility.NONE
)
@JsonPropertyOrder({"lang", "name", "value"})
public abstract class V2ArgumentMixin {

  @JsonProperty("name")
  private String name;

  @JsonSerialize(using = LangCustomSerializer.class, as=String.class)
  @JsonProperty("lang")
  private String lang;

  @JsonProperty("value")
  private Object value;


  
  public static class LangCustomSerializer extends JsonSerializer<String> {

    @Override
    public void serialize(String value,
                          JsonGenerator jsonGenerator,
                          SerializerProvider serializerProvider)
        throws IOException, JsonProcessingException {
      jsonGenerator.writeObject(value.toString() + "  - something else");
    }
  }
}
于 2020-12-17T18:47:04.043 回答