java - 从 xml 文件检索值的 JAXB 程序

Question

我正在尝试使用 JAXB 从 xml 检索值。下面详细介绍一下:

我的 OrderValidation.xml

 <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
    <OrderValidation>
        <OrderType name="Activation-Activation">
            <Product>
                <Name>TV_SAT</Name>
                <ActionCode>ADD</ActionCode>
            </Product>
        </OrderType>
        <OrderType name="Change Owner-Change Owner">
            <Product>
                <Name>TV_SAT_EQUIPMENT</Name>
                <ActionCode>EXISTING</ActionCode>
            </Product>
        </OrderType>
    </OrderValidation>

这是我的 Orderxml.java

@XmlRootElement(name ="OrderValidation")
public class Orderxml 
{
    private String name;
    private String Product;
    private String OrderType;

    public Orderxml() {}
    public Orderxml(String name, String productclass, String ordertype) 
    {
        super();

        this.name = name;
        this.Product = productclass;
        this.OrderType = ordertype;
    }
    @XmlAttribute
    public String getName()
    {
        return name;
    }
    public void setName(String name)
    {
        this.name = name;
    }
    @XmlElement
    public String getProductName()
    {
        return Product;
    }
    public void setProductName(String productclass)
    {
        this.Product = productclass;
    }
    @XmlElement
    public String getOrderType()
    {
        return OrderType;
    }
    public void setOrderType(String ordertype)
    {
        this.OrderType = ordertype;
    }
}

我的UnMarshall代码(主类)

public static void main(String[] args)
{
     try
     {
        File file = new File("C:///OrderValidation.xml");
        System.out.println(1);
        JAXBContext jaxbContext = JAXBContext.newInstance(Orderxml.class);
        System.out.println(2);
        Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
        System.out.println(3);
        Orderxml ord= (Orderxml) jaxbUnmarshaller.unmarshal(file);
        System.out.println(4);
        System.out.println(ord.getOrderType()+". "+ord.getProductName());
     }
     catch (JAXBException e)
     {
       System.out.println("Issue is here");
       e.printStackTrace();
     }
}

当我执行代码时，它会抛出 null 作为输出。请建议我更改代码以便检索值。提前致谢

Answer 1

你们的 JAXB POJO 似乎不正确。根据您的 XML 结构，它应该类似于:

@XmlRootElement(name ="OrderValidation")
public class Orderxml 
{
    @XmlElement("OrderType") 
    private List<OrderType> orderTypes;
}



public class OrderType{
    @XmlElement("Product")
    private Product product;
    @XmlAttribute(name="name")
    private String name;
}


public class Product {
  @XmlElement("Name")
  private String name;
  @XmlElement("ActionCode")
  private String actionCode;
}

您可以使用在线工具从 XML 生成近似的 xsd，然后使用 schemagen 实用程序生成 POJO，而不是手写它。