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，我可以得到描述和外部密钥。但是我不能得到有价值的属性名称。

>

这是我的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？谢谢你

Answer 1

银行

您需要将属性属性从String更改为域对象。

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

所有物

然后，您的属性对象看起来像：

@XmlAccessorType(XmlAccessType.FIELD)
public class Property {

    @XmlAttribute
    private String name;

    @XmlAtrribute
    private String value;

}