Question

我使用JAXB注释来使用Spring序列化/反序列化对象。

这是我的回复课程：

@XmlRootElement(name = "animals")
public class PetClinic implements Serializable {

    @XmlElementWrapper(name = "animals")
    @XmlElement(name = "animal")
    private Collection<Animal> animals;

    public Collection<Animal> getAnimals() {
        return animals;
    }

    public void setAnimalsCollection<Animal> animals) {
        this.animals = animals;
    }

}

我想获得像这样的JSON

{
    "animals": [
        {...}, 
        {...}
    ]
}

我试图解开根元素，但我的所有解决方案都不起作用：

1）我试图在字段

上使用@JsonUnwrapped注释

2）我尝试使用@JsonSerialize

@JsonSerialize(using = PetClinicSerializer.class)

串行器代码：

ObjectMapper mapper = new ObjectMapper();

@Override
public void serialize(PetClinic clinic, JsonGenerator generator, SerializerProvider serializerProvider) throws IOException, JsonProcessingException {
    mapper.configure(SerializationFeature.WRAP_ROOT_VALUE, false);
    String clinicAsString = mapper.writeValueAsString(clinic);
    generator.writeString(clinicAsString);
}

如何获得上述JSON？

Answer 1

我刚刚使用ObjectMapper而没有任何jaxb表示法并获得了所需的结果：

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import java.io.Serializable;
import java.util.Arrays;
import java.util.Collection;

public class PetClinic implements Serializable {

    private Collection<Animal> animals;

    public Collection<Animal> getAnimals() {
        return animals;
    }

    public void setAnimals(Collection<Animal> animals) {
        this.animals = animals;
    }

    public static void main(String[] args) throws Exception {
        PetClinic pc = new PetClinic();
        Animal cat = new Animal();
        cat.setName("cat");
        Animal dog = new Animal();
        dog.setName("dog");
        pc.setAnimals(Arrays.asList(cat, dog));
        ObjectMapper mapper = new ObjectMapper();
        mapper.enable(SerializationFeature.INDENT_OUTPUT);
        String s = mapper.writeValueAsString(pc);
        System.out.println(s);

    }

}

class Animal {

    private String name;

    public String getName() {
        return name;
    }

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

输出

{
  "animals" : [ {
    "name" : "cat"
  }, {
    "name" : "dog"
  } ]
}

使用JAXB注释时如何为JSON解包？

1 个答案: