Jackson 如何将 json 反序列化为泛型类型？

Question

类人：

@Data
public class Person<T extends Parent> implements Serializable {
    private static final long serialVersionUID = 7822965786010192978L;

    private static final ObjectMapper objectMapper = new ObjectMapper();  
    private String id;    
    private T people; // change (String peopleInfo) to object extends Parent after read data from database
    private String peopleInfo; // change (T people) to string and save in the dataBase as an string

    @SneakyThrows
    public void setPeople(T people) {
        this.people = people;
        peopleInfo = objectMapper.writeValueAsString(people);
    }

    @SneakyThrows
    public void setPeopleInfo(String peopleInfo) {
        this.peopleInfo = peopleInfo;
        if (!Strings.isNullOrEmpty(peopleInfo)) {
            people = objectMapper.readValue(peopleInfo, new TypeReference<T>() {});
        }
    }
}

班级家长：

@Data
public class Parent implements Serializable {
    private static final long serialVersionUID = 2092353331893381153L;
    private String name;
}

类孩子：

@Data
public class Child extends Parent {
    private static final long serialVersionUID = 3318503314202792570L;
    private String pocketMoney;
}

和测试功能：我想将persnInfo保存到数据库中为String，并在从数据库中读取数据后自动将string更改为对象people。

@Test
public void testReadData() throws Exception {
    Child child = new Child();
    child.setName("_child");
    child.setPocketMoney("10$");

    Person<Child> person = new Person<>();
    person.setId("1");
    person.setPeople(child);

    // assume this json was read from database
    String json = person.getPeopleInfo();
    System.out.println(json);

    Person<Child> readPerson = new Person<>();
    readPerson.setId("1");
    readPerson.setPeopleInfo(json);

    Child readChild = readPerson.getPeople();
    System.out.println(readChild.getPocketMoney());
}

发生错误：

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException：无法识别的字段“pocketMoney”（类pers.test.common.objmapper.Parent），未标记为可忽略（一个已知属性：“名称”]）

在 [来源：{"name":"_child","pocketMoney":"10$"}; line: 1, column: 33]（通过引用链：pers.test.common.objmapper.Parent["pocketMoney"]）

在 com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:51)

我不知道如何解决这个问题，任何人都可以在不更改 testReadData()' but change the modelPeople` 的情况下帮助我让它通过。

Answer 1

你的问题在这里：

new TypeReference<T>()

这并不符合您的预期。 Java 泛型在运行时被擦除；因此上面的语句基本上是new TypeReference<Object>。

换句话说 - 你声明的事实

Person<Child> readPerson = new Person<>();

到 expect 子对象根本不够！

您可能必须将特定类 Child.class 传递给将 JSON 字符串映射回的代码。如需更多信息，请查看here。