Java：如何创建参数化的Jackson Json到对象函数

Question

I am trying to serialize objects to and deserialize from json. I tried to make a class with a static method that would accomplish this in the general case given a class.

public static <T> T jsonToObject(String json, Class<T> tClass) {
    try {
        return mapper.readValue(json, tClass);
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

Then in a unit test:

@Test
public void testUserDataSerialization() {

    UserData userData = new UserData
        .Builder("david@davidgeorgewilliams.com")
        .firstName("david")
        .lastName("Williams")
        .build();

    String userDataJson = JsonUtils.objectToJson(userData);

    String userDataJson1 = JsonUtils.objectToJson(userData1);

    System.out.println(userDataJson + "\n" + userData1);
} 

This blows up with this exception

Running com.company.app.UserDataJsonTest
org.codehaus.jackson.map.JsonMappingException: No suitable constructor found for type [simple type, class com.company.app.model.UserData]: can not instantiate from JSON object (need to add/enable type information?)
at [Source: java.io.StringReader@76ba6b2; line: 1, column: 2]
at org.codehaus.jackson.map.JsonMappingException.from(JsonMappingException.java:163)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObjectUsingNonDefault(BeanDeserializer.java:746)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:683)

What am I doing wrong here and how do I fix it? The idea is I want to accept JSON from a post request body, and turn it into the corresponding object.

Answer 1

You need to use TypeReference to handle type erasure with generics. Check out the documentation (the Data Binding with Generics section). TypeFactory works as well.

Answer 2

答案如上所述。