Question

我想在JavaSE独立应用程序中使用JAXB来读写Json文件。我设法使用下面的代码片段为XML文件执行此操作，但我不明白如何切换到Json：

public class Main {

  public static void main(String[] args) throws Exception {
    Book book = new Book();
    book.title = "hello";

    JAXBContext context = JAXBContext.newInstance(Book.class);
    Marshaller marshaller = context.createMarshaller();
    marshaller.marshal(book, System.out);
  }

}

注意：

表现不是问题。
Web服务/ REST不在话题中。

Answer 1

使用EclipseLink MOXy，您可以添加属性来操作输出。

    public class Main {

    public static void main(String[] args) {
        Book book = new Book();
        book.title = "hello";

        JAXBContext context;
        try {
            context = JAXBContextFactory.createContext(new Class[] {Book.class}, null);
            Marshaller marshaller = context.createMarshaller();
            marshaller.setProperty(MarshallerProperties.MEDIA_TYPE, "application/json");
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            marshaller.marshal(book, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }

}

会导致： { “书”：{ “标题”：“你好” } }

类路径上需要org.eclipse.persistence.moxy-2.5.1.jar和org.eclipse.persistence.core-2.5.1.jar。在我自己玩这个时，我遇到了：hottest JAXB answers。特别是Blaise Doughan在非常有帮助的地方回答。搜索

MarshallerProperties.MEDIA_TYPE, "application/json"

更多他的例子。

Answer 2

如果你愿意，你可以创建一个单独的方法就像这样非常简单（如果你已经有了JAXB库）

主要班级

import java.io.IOException;
import javax.xml.bind.JAXBException;
import org.codehaus.jackson.JsonParseException;
import org.codehaus.jackson.map.JsonMappingException;
import org.codehaus.jackson.map.ObjectMapper;

public class ConvertJson {

    public static void main(String[] args) {
        final JavaObject javaObject = new JavaObject();
        javaObject.setName("Json");
        javaObject.setRole("Moderator");
        javaObject.setAge(28);

        try {
            pojo2Json(javaObject);
        } catch (JAXBException | IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

    private static String pojo2Json(Object obj) throws JAXBException,
            JsonParseException, JsonMappingException, IOException {
        ObjectMapper objectMapper = new ObjectMapper();
        String jsonString = objectMapper.writeValueAsString(obj);
        System.out.print(jsonString);
        return jsonString;
    }

这是pojo

import org.codehaus.jackson.annotate.JsonProperty;
public class JavaObject{
    @JsonProperty(value = "Name")
    private String name;
    @JsonProperty(value = "Role")
    private String role;
    @JsonProperty(value = "Age")
    private int age;
    public JavaObject(){
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getRole() {
        return role;
    }
    public void setRole(String role) {
        this.role = role;
    }
    public int getAge() {
        return age;
    }
    public void setAge(int age) {
        this.age = age;
    }

}

}

Answer 3

你会这样做：

public class Main {
 public static void main(String[] args) throws IOException {
  Book book = new Book();
  book.title = "hello";

  ObjectMapper mapper = new ObjectMapper();
  mapper.writeValue(System.out, book);
 }
}

在独立JavaSE应用程序中与JAXB和Jackson的JSON绑定

3 个答案: