31

我在我的项目中使用 FasterXML/Jackson-Databind 已经有一段时间了，一切都很好，直到我发现了这篇文章并开始使用这种方法来反序列化没有 @JsonProperty 注释的对象。

问题是，当我有一个带有多个参数的构造函数并用 @JsonCreator 注释装饰这个构造函数时，杰克逊会抛出以下错误：

Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: 
Argument #0 of constructor [constructor for com.eliti.model.Cruiser, annotations: {interface com.fasterxml.jackson.annotation.JsonCreator=@com.fasterxml.jackson.annotation.JsonCreator(mode=DEFAULT)}] has no property name annotation; must have name when multiple-parameter constructor annotated as Creator
 at [Source: {
  "class" : "com.eliti.model.Cruiser",
  "inventor" : "afoaisf",
  "type" : "MeansTransport",
  "capacity" : 123,
  "maxSpeed" : 100
}; line: 1, column: 1]

我创建了一个小项目来说明这个问题，我试图反序列化的类是这个：

public class Cruise extends WaterVehicle {

 private Integer maxSpeed;

  @JsonCreator
  public Cruise(String name, Integer maxSpeed) {
    super(name);
    System.out.println("Cruise.Cruise");
    this.maxSpeed = maxSpeed;
  }

  public Integer getMaxSpeed() {
    return maxSpeed;
  }

  public void setMaxSpeed(Integer maxSpeed) {
    this.maxSpeed = maxSpeed;
  }

}

反序列化的代码是这样的：

public class Test {
  public static void main(String[] args) throws IOException {
    Cruise cruise = new Cruise("asd", 100);
    cruise.setMaxSpeed(100);
    cruise.setCapacity(123);
    cruise.setInventor("afoaisf");

    ObjectMapper mapper = new ObjectMapper().enable(SerializationFeature.INDENT_OUTPUT);
    mapper.registerModule(new ParameterNamesModule(JsonCreator.Mode.PROPERTIES));

    String cruiseJson = mapper.writeValueAsString(cruise);

    System.out.println(cruiseJson);

    System.out.println(mapper.readValue(cruiseJson, Cruise.class));

}

我已经尝试删除 @JsonCreator，但如果这样做，则会引发以下异常：

Exception in thread "main" com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of com.eliti.model.Cruise: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?)
 at [Source: {
  "class" : "com.eliti.model.Cruise",
  "inventor" : "afoaisf",
  "type" : "MeansTransport",
  "capacity" : 123,
  "maxSpeed" : 100
}; line: 3, column: 3]

我试图发出“mvn clean install”，但问题仍然存在。

为了包含一些额外的信息，我已经彻底研究了这个问题（GitHub 问题、博客文章、StackOverflow 问答）。以下是我一直在做的一些调试/调查：

调查一

生成的字节码上的javap -v 给我这个：

 MethodParameters:
      Name                           Flags
      name
      maxSpeed

在谈论构造函数时，我猜想-parameters标志确实是为 javac 编译器设置的。

调查二

如果我创建一个带有单个参数的构造函数，则对象会被初始化，但我想要/需要使用多参数构造函数。

调查3

如果我在每个字段上使用注释@JsonProperty，它也可以工作，但是对于我的原始项目来说，开销太大，因为我在构造函数中有很多字段（而且使用注释重构代码也非常困难）。

剩下的问题是： 如何让 Jackson 在没有注释的情况下使用多参数构造函数？

4

4 回答 4

41

您需要添加注解，@JsonProperty指定创建对象时需要传递给构造函数的 json 属性的名称。

public class Cruise extends WaterVehicle {

 private Integer maxSpeed;

  @JsonCreator
  public Cruise(@JsonProperty("name") String name, @JsonProperty("maxSpeed")Integer maxSpeed) {
    super(name);
    System.out.println("Cruise.Cruise");
    this.maxSpeed = maxSpeed;
  }

  public Integer getMaxSpeed() {
    return maxSpeed;
  }

  public void setMaxSpeed(Integer maxSpeed) {
    this.maxSpeed = maxSpeed;
  }

}

编辑

我刚刚使用下面的代码进行了测试，它对我有用

import java.io.IOException;

import com.fasterxml.jackson.annotation.JsonCreator.Mode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.module.paramnames.ParameterNamesModule;

class WaterVehicle {

    private String name;
    private int capacity;
    private String inventor;
    public WaterVehicle(String name) {
        this.name=name;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public int getCapacity() {
        return capacity;
    }
    public void setCapacity(int capacity) {
        this.capacity = capacity;
    }
    public String getInventor() {
        return inventor;
    }
    public void setInventor(String inventor) {
        this.inventor = inventor;
    }


}

 class Cruise  extends WaterVehicle{

        private Integer maxSpeed;

        public Cruise(String name, Integer maxSpeed) {
            super(name);
            this.maxSpeed = maxSpeed;
        }

        public Integer getMaxSpeed() {
            return maxSpeed;
        }

        public void setMaxSpeed(Integer maxSpeed) {
            this.maxSpeed = maxSpeed;
        }


    }

public class Test {
      public static void main(String[] args) throws IOException {
        Cruise cruise = new Cruise("asd", 100);
        cruise.setMaxSpeed(100);
        cruise.setCapacity(123);
        cruise.setInventor("afoaisf");

        ObjectMapper mapper = new ObjectMapper().enable(SerializationFeature.INDENT_OUTPUT);
        mapper.registerModule(new ParameterNamesModule(Mode.PROPERTIES));

        String jsonString = mapper.writeValueAsString( cruise);
        System.out.println(jsonString);

        Cruise anotherCruise = mapper.readValue(jsonString, Cruise.class);
         System.out.println(anotherCruise );
         jsonString = mapper.writeValueAsString( anotherCruise );
         System.out.println(jsonString);

    }

}

它产生以下输出

{
  "name" : "asd",
  "capacity" : 123,
  "inventor" : "afoaisf",
  "maxSpeed" : 100
}
Cruise@56f4468b
{
  "name" : "asd",
  "capacity" : 123,
  "inventor" : "afoaisf",
  "maxSpeed" : 100
}

确保 pom 文件中有 compilerArgs。

<compilerArgs>
     <arg>-parameters</arg>
</compilerArgs>
于 2016-08-24T12:26:32.230 回答
12

简短回答：使用 Java 8javac -parameters和jackson-module-parameter-names

长答案：为什么当构造函数使用 @JsonCreator 注释时，它的参数必须使用 @JsonProperty 注释？

于 2016-08-24T13:22:00.840 回答
8

@JsonCreator@JsonProperty("xxx")在参数中有后不需要

于 2019-01-07T05:21:47.697 回答
0

如果您有嵌套的内部类，请将其设为静态

public class A{
    private int property1;
    public static class B{
        private String property2;
    }
}
于 2021-09-29T07:53:39.957 回答