Question

我正在尝试将json api响应映射到一个对象，而IntelliJ正在抱怨。它在说cannot resolve method readValue(java.lang.String, java.lang.Object[]);。我意识到我没有传递正确的参数，但我已经尝试了responseClass.class和responseClass.getClass()而没有运气。

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

这是使用Jackson将JSON响应映射到对象的一种非常奇怪的方式。我假设您只想使用实用程序类映射任意类？这是一个更容易实现这一目标的例子。请注意，这使用了杰克逊发行版附带的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，仍然可以使用您发布的课程从JSON映射您的回复，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
    }
}

并使用它：

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

如何将正确的对象传递给mapper.readValue（）

1 个答案: