如何使用JAXB获取XML属性

Question

如何使用JAXB获取XML属性

6

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>

我使用了JAXB和unmarshall来处理这个XML，可以获取到description和externalkey。但是我无法获取带有值的属性名称。

This is my java class for unmarshall:

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());

Organization.java

@XmlRootElement
public class Organization {
Bank bank = new Bank();

public Bank getBank() {
  return bank;
}

public void setBank(Bank bank) {
 this.bank = bank;
}
}

Bank.java

@XmlRootElement
public class Bank {
 private String description;
 private String externalkey;
 private String property;

//..GETTER and SETTER
}

How can I get property name and value? Thank u

- kamal

1

展示你的Organization类。 - BobTheBuilder

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- bdoughan · Accepted Answer

银行

您需要将 property 属性从一个 String 类型更改为域对象。

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

属性

那么您的属性对象可能看起来像这样：

@XmlAccessorType(XmlAccessType.FIELD)
public class Property {

    @XmlAttribute
    private String name;

    @XmlAtrribute
    private String value;

}