Question

我有一个下面的Enum desrcibed：

public enum OrderType {

  UNKNOWN(0, "Undefined"),
  TYPEA(1, "Type A"),
  TYPEB(2, "Type B"),
  TYPEC(3, "Type C");

  private Integer id;
  private String name;

  private WorkOrderType(Integer id, String name) {
    this.id = id;
    this.name = name;
  }

  //Setters, getters....
}

我使用我的控制器（new OrderType[] {UNKNOWN,TYPEA,TYPEB,TYPEC};）返回枚举数组，然后Spring将其序列化为以下json字符串：

["UNKNOWN", "TYPEA", "TYPEB", "TYPEC"]

迫使杰克逊像POJO一样序列化枚举的最佳方法是什么？ E.g：

[
  {"id": 1, "name": "Undefined"},
  {"id": 2, "name": "Type A"},
  {"id": 3, "name": "Type B"},
  {"id": 4, "name": "Type C"}
]

我玩了不同的注释，却无法获得这样的结果。

Answer 1

最后我自己找到了解决方案。

我必须使用@JsonSerialize(using = OrderTypeSerializer.class)注释枚举并实现自定义序列化程序：

public class OrderTypeSerializer extends JsonSerializer<OrderType> {

  @Override
  public void serialize(OrderType value, JsonGenerator generator,
            SerializerProvider provider) throws IOException,
            JsonProcessingException {

    generator.writeStartObject();
    generator.writeFieldName("id");
    generator.writeNumber(value.getId());
    generator.writeFieldName("name");
    generator.writeString(value.getName());
    generator.writeEndObject();
  }
}

Answer 2

@JsonFormat(shape= JsonFormat.Shape.OBJECT)
public enum SomeEnum

自https://github.com/FasterXML/jackson-databind/issues/24

以来可用

刚测试它适用于版本2.1.2

回答TheZuck ：

我试过你的例子，得到了Json：

{"events":[{"type":"ADMIN"}]}

我的代码：

@RequestMapping(value = "/getEvent") @ResponseBody
  public EventContainer getEvent() {
    EventContainer cont = new EventContainer();
    cont.setEvents(Event.values());
    return cont;
 }

class EventContainer implements Serializable {

  private Event[] events;

  public Event[] getEvents() {
    return events;
 }

 public void setEvents(Event[] events) {
   this.events = events;
 }
}

和依赖关系是：

<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-annotations</artifactId>
  <version>${jackson.version}</version>
</dependency>

<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-core</artifactId>
  <version>${jackson.version}</version>
</dependency>

<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-databind</artifactId>
  <version>${jackson.version}</version>
  <exclusions>
    <exclusion>
      <artifactId>jackson-annotations</artifactId>
      <groupId>com.fasterxml.jackson.core</groupId>
    </exclusion>
    <exclusion>
      <artifactId>jackson-core</artifactId>
      <groupId>com.fasterxml.jackson.core</groupId>
    </exclusion>
  </exclusions>
</dependency>

<jackson.version>2.1.2</jackson.version>

Answer 3

我找到了一个非常简洁的解决方案，当你无法像我的情况那样修改枚举类时尤其有用。然后，您应该提供自定义的ObjectMapper，并启用某个功能。自Jackson 1.6以来就可以使用这些功能。

public class CustomObjectMapper extends ObjectMapper {
    @PostConstruct
    public void customConfiguration() {
        // Uses Enum.toString() for serialization of an Enum
        this.enable(WRITE_ENUMS_USING_TO_STRING);
        // Uses Enum.toString() for deserialization of an Enum
        this.enable(READ_ENUMS_USING_TO_STRING);
    }
}

有更多与枚举相关的功能，请参见此处：

https://github.com/FasterXML/jackson-databind/wiki/Serialization-features https://github.com/FasterXML/jackson-databind/wiki/Deserialization-Features

Answer 4

这是我的解决方案。我想将变换枚举转换为{id: ..., name: ...}形式。

使用Jackson 1.x ：

的pom.xml：

<properties>
    <jackson.version>1.9.13</jackson.version>
</properties>

<dependencies>
    <dependency>
        <groupId>org.codehaus.jackson</groupId>
        <artifactId>jackson-core-asl</artifactId>
        <version>${jackson.version}</version>
    </dependency>
    <dependency>
        <groupId>org.codehaus.jackson</groupId>
        <artifactId>jackson-mapper-asl</artifactId>
        <version>${jackson.version}</version>
    </dependency>
</dependencies>

Rule.java：

import org.codehaus.jackson.map.annotate.JsonSerialize;
import my.NamedEnumJsonSerializer;
import my.NamedEnum;

@Entity
@Table(name = "RULE")
public class Rule {
    @Column(name = "STATUS", nullable = false, updatable = true)
    @Enumerated(EnumType.STRING)
    @JsonSerialize(using = NamedEnumJsonSerializer.class)
    private Status status;
    public Status getStatus() { return status; }
    public void setStatus(Status status) { this.status = status; }

    public static enum Status implements NamedEnum {
        OPEN("open rule"),
        CLOSED("closed rule"),
        WORKING("rule in work");

        private String name;
        Status(String name) { this.name = name; }
        public String getName() { return this.name; }
    };
}

NamedEnum.java：

package my;

public interface NamedEnum {
    String name();
    String getName();
}

NamedEnumJsonSerializer.java：

package my;

import my.NamedEnum;
import java.io.IOException;
import java.util.*;
import org.codehaus.jackson.JsonGenerator;
import org.codehaus.jackson.JsonProcessingException;
import org.codehaus.jackson.map.JsonSerializer;
import org.codehaus.jackson.map.SerializerProvider;

public class NamedEnumJsonSerializer extends JsonSerializer<NamedEnum> {
    @Override
    public void serialize(NamedEnum value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
        Map<String, String> map = new HashMap<>();
        map.put("id", value.name());
        map.put("name", value.getName());
        jgen.writeObject(map);
    }
}

使用Jackson 2.x ：

的pom.xml：

<properties>
    <jackson.version>2.3.3</jackson.version>
</properties>

<dependencies>
    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-core</artifactId>
        <version>${jackson.version}</version>
    </dependency>
    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-databind</artifactId>
        <version>${jackson.version}</version>
    </dependency>
</dependencies>

Rule.java：

import com.fasterxml.jackson.annotation.JsonFormat;

@Entity
@Table(name = "RULE")
public class Rule {
    @Column(name = "STATUS", nullable = false, updatable = true)
    @Enumerated(EnumType.STRING)
    private Status status;
    public Status getStatus() { return status; }
    public void setStatus(Status status) { this.status = status; }

    @JsonFormat(shape = JsonFormat.Shape.OBJECT)
    public static enum Status {
        OPEN("open rule"),
        CLOSED("closed rule"),
        WORKING("rule in work");

        private String name;
        Status(String name) { this.name = name; }
        public String getName() { return this.name; }
        public String getId() { return this.name(); }
    };
}

Rule.Status.CLOSED已翻译为{id: "CLOSED", name: "closed rule"}。

Answer 5

序列化Enum的一种简单方法是使用@JsonFormat注释。 @JsonFormat可以通过三种方式配置Enum的序列化。

@JsonFormat.Shape.STRING
public Enum OrderType {...}

使用OrderType :: name作为序列化方法。 OrderType.TypeA的序列化为“TYPEA”

@JsonFormat.Shape.NUMBER
Public Enum OrderTYpe{...}

使用OrderType :: ordinal作为序列化方法。 OrderType.TypeA的序列化为1

@JsonFormat.Shape.OBJECT
Public Enum OrderType{...}

将OrderType视为POJO。 OrderType.TypeA的序列化为{"id":1,"name":"Type A"}

你需要

JsonFormat.Shape.OBJECT 。

更复杂的方法是你的解决方案，为Enum指定一个序列化器。

查看此参考： https://fasterxml.github.io/jackson-annotations/javadoc/2.2.0/com/fasterxml/jackson/annotation/JsonFormat.html

Answer 6

使用@JsonCreator注释，创建方法getType（），使用toString或对象工作进行序列化

{"ATIVO"}

或

{"type": "ATIVO", "descricao": "Ativo"}

...

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.node.JsonNodeType;

@JsonFormat(shape = JsonFormat.Shape.OBJECT)
public enum SituacaoUsuario {

    ATIVO("Ativo"),
    PENDENTE_VALIDACAO("Pendente de Validação"),
    INATIVO("Inativo"),
    BLOQUEADO("Bloqueado"),
    /**
     * Usuarios cadastrados pelos clientes que não possuem acesso a aplicacao,
     * caso venham a se cadastrar este status deve ser alterado
     */
    NAO_REGISTRADO("Não Registrado");

    private SituacaoUsuario(String descricao) {
        this.descricao = descricao;
    }

    private String descricao;

    public String getDescricao() {
        return descricao;
    }

    // TODO - Adicionar metodos dinamicamente
    public String getType() {
        return this.toString();
    }

    public String getPropertieKey() {
        StringBuilder sb = new StringBuilder("enum.");
        sb.append(this.getClass().getName()).append(".");
        sb.append(toString());
        return sb.toString().toLowerCase();
    }

    @JsonCreator
    public static SituacaoUsuario fromObject(JsonNode node) {
        String type = null;
        if (node.getNodeType().equals(JsonNodeType.STRING)) {
            type = node.asText();
        } else {
            if (!node.has("type")) {
                throw new IllegalArgumentException();
            }
            type = node.get("type").asText();
        }
        return valueOf(type);
    }

}

Answer 7

在Spring Boot 2中，最简单的方法是在application.properties中声明：

spring.jackson.serialization.WRITE_ENUMS_USING_TO_STRING=true
spring.jackson.deserialization.READ_ENUMS_USING_TO_STRING=true

并定义枚举的toString（）方法。

用Jackson序列化枚举

7 个答案: