Question

我有一个JSON文件，其中包含我需要用于创建对象的2个值。像这样：

{ 
    "name": ["Adam", "Bart"],
    "surname": ["White","Brown"]
}

我想使用这两个属性来创建一个包含Person（name，surname）类型的两个对象的List。

我如何使用杰克逊实现这一目标？

Answer 1

您可以使用自定义反序列化器实现此目的：

我使用了这个Person类：

public class Person
{
    public String name;
    public String surName;
    public Person (String name, String surName) {
        this.name = name;
        this.surName = surName;
    }
    public String toString() {
        return name + " " + surName;
    }
}

这是自定义的Json反序列化器：

@SuppressWarnings("serial")
public class PersonListDeserializer extends StdDeserializer<List<Person>>
{
    public PersonListDeserializer() { 
        this(null);
    } 

    public PersonListDeserializer(Class<?> vc) { 
        super(vc); 
    }

    @Override
    public List<Person> deserialize(JsonParser jp, DeserializationContext ctxt) 
            throws IOException, JsonProcessingException
    {
        List<Person> returnList = new ArrayList<>();
        JsonNode rootNode = jp.getCodec().readTree(jp);
        // expecting to find to arrays under root:
        ArrayNode nameNode = (ArrayNode)rootNode.get("name");
        ArrayNode surnameNode = (ArrayNode)rootNode.get("surname");
        // build the return list from arrays
        for (int i = 0 ; i < nameNode.size() ; i++) {
            returnList.add(new Person(nameNode.get(i).toString(), surnameNode.get(i).toString()));
        }
        return returnList;
    }
}

全部放在一起：

public static void main(String[] args)
{
    String jsonString = "{ \"name\": [\"Adam\", \"Bart\"], \"surname\": [\"White\",\"Brown\"] }";

    SimpleModule module = new SimpleModule();
    // Note: the custom deserializer is registered to invoke for every List in input! 
    module.addDeserializer(List.class, new PersonListDeserializer());
    ObjectMapper mapper = new ObjectMapper();
    mapper.registerModule(module);

    try {
        // instructing jackson of target generic list type 
        CollectionType personListType = TypeFactory.defaultInstance().constructCollectionType(List.class, Person.class);
        @SuppressWarnings("unchecked")
        List<Person> personList = (List<Person>)mapper.readValue(jsonString, personListType);
        System.out.println(personList);
    } catch (IOException e) {
        e.printStackTrace();
    }
}

输出符合预期：

["Adam" "White", "Bart" "Brown"]

使用两个或多个属性创建一个对象属性

1 个答案: