java - 如何将正确的对象传递给mapper.readValue()

Question

我正在尝试将 json api 响应映射到对象，而 IntelliJ 正在提示。它说无法解析方法readValue(java.lang.String, java.lang.Object[]);。我意识到我没有传递正确的参数，但我尝试了 responseClass.class 和 responseClass.getClass() 但没有成功。

用法:

MyClass myClass = new MyClass();
myClass.setResponseClass(User.class);

定义:

MyClass {
    private Object responseClass;

    public void setResponseClass(Object responseClass) {
        this.responseClass = responseClass;
    }

    public Object getResponseClass() {
        return responseClass;
    }

    public void getApiResponse() {
        //some code here

        ObjectMapper mapper = new com.MyApp.Utility.ObjectMapper();

        //some code here

        //I've tried responseClass.class and responseClass.getClass(), it didn't like either of them
        mapper.readValue(response, responseClass);

        //more code here
    }
}

Answer 1

这是一种使用 Jackson 将 JSON 响应映射到对象的非常奇怪的方法。我假设您只是希望能够使用实用程序类映射任意类？这是实现此目的的更简单方法的示例。请注意，这使用 Jackson 发行版附带的 ObjectMapper 类:

public class JSONUtil {
    private ObjectMapper mapper = new ObjectMapper();

    public JSONUtil() {
        super();
        // Set ObjectMapper configuration and properties here
    }

    public <T> T deserialize(final String response, final Class<T> responseClass) {
        if(response == null || responseClass == null) return null;

        return mapper.readValue(response, responseClass);
    }
}

现在，您仍然可以使用您发布的类映射来自 JSON 的响应，并进行一些修改:

public class MyClass<T> {
    private Class<T> responseClass;

    public MyClass(final Class<T> responseClass) {
        super();
        this.responseClass = responseClass;
    }

    public void getApiResponse(final String response) {

        final ObjectMapper mapper = new ObjectMapper();
        final T values = mapper.readValue(response, responseClass);

        //more code here
    }
}

并这样使用它:

MyClass<User> myClass = new MyClass<User>(User.class);
myClass.getApiResponse(someJsonString);