java - 如何使用 JAXB 获取 xml 属性

Question

这是我的xml：

<?xml version="1.0" encoding="UTF-8" ?>
    <organization>
      <bank>
        <description>aaa</description>
        <externalkey>123</externalkey>
        <property name="pName" value="1234567890" />
      </bank>
   </organization>

我为这个xml使用了JAXB和unmarshall，我可以获得描述和externalkey。但我无法获得具有价值的属性名称。

• 这是我用于解组的 java 类：

  JAXBContext jb = JAXBContext.newInstance(Organization.class);
  Unmarshaller um = jb.createUnmarshaller();
  Organization org = (Organization) um.unmarshal(new File("\\upload\\bank999999.xml"));
  System.out.println(org.getBank().getDescription());
  System.out.println(org.getBank().getExternalkey());
  

• 组织.java

  @XmlRootElement
  public class Organization {
  Bank bank = new Bank();
  
  public Bank getBank() {
    return bank;
  }
  
  public void setBank(Bank bank) {
   this.bank = bank;
  }
  }
  

• 银行.java

  @XmlRootElement
  public class Bank {
   private String description;
   private String externalkey;
   private String property;
  
  //..GETTER and SETTER
  }
  

  如何获取属性名称和值？感谢你

Answer 1

银行

您需要将property属性从 a更改String为域对象。

@XmlAccessorType(XmlAccessType.FIELD)
public class Bank {
    private String description;
    private String externalkey;
    private Property property;
}

财产

然后你的Property对象看起来像：

@XmlAccessorType(XmlAccessType.FIELD)
public class Property {

    @XmlAttribute
    private String name;

    @XmlAtrribute
    private String value;

}