如何为不同的类写一个接口实现？

Question

我想为不同类型的类编写一个实现。
这是 interface:

public interface ResourcesInterface<T> {
  T readJsonContent(String fileName/*, maybe there also must be class type?*/);
}

这是 interface Student.class 的实现。在下面的示例中，我尝试读取 JSON 文件并从中接收 Student.class 对象：

import com.fasterxml.jackson.databind.ObjectMapper;

public class StudentResources implements ResourcesInterface<Student> {

  @Override
  public Student readJsonContent(String fileName) {
    Student student = new Student();
    ObjectMapper objectMapper = new ObjectMapper();

    try {
      URL path = getClass().getClassLoader().getResource(fileName);
      if (path == null) throw new NullPointerException();
      student = objectMapper.readValue(path, Student.class);

    } catch (IOException exception) {
      exception.printStackTrace();
    }

    return student;
  }
}

所以我不想为每个 class 类型实现这个 interface 我想使用方法 readJsonContent(String) 像这样：

Student student = readFromJson(fileName, Student.class);
AnotherObject object = readFromJson(fileName, AnotherObject.class);

有没有可能只写一个实现？而不是为每个不同的 class 多次实施 interface？任何想法如何做到这一点？

Answer 1

如果我理解正确的话，您需要一个能够将 JSON 文件解码为对象的通用方法，对吗？如果是这样，那么您不需要接口。您只需要使用如下静态方法创建一个 class：

import org.codehaus.jackson.map.ObjectMapper;

import java.io.IOException;
import java.io.UncheckedIOException;
import java.net.URL;
import java.util.Objects;

public class JsonUtil  {

    private JsonUtil(){}

    public static <T> T readJsonContent(String fileName, Class<T> clazz) {
        ObjectMapper objectMapper = new ObjectMapper();
        try {
            URL path = Objects.requireNonNull(clazz.getResource(fileName));
            return objectMapper.readValue(path, clazz);
        } catch (IOException ex) {
            throw new UncheckedIOException("Json decoding error", ex);
        }
    }

    public static void main(String[] args) {
        Student s = JsonUtil.readJsonContent("", Student.class);
    }
}

如何为不同的类写一个接口实现？

How to write one interface implementation for different classes?

java

abstract-class

interface