Java Unmarshal带有JAXB的类包装器的对象列表

Question

从BaseX服务器执行的XQuery我得到一个结果:

<ProtocolloList>
  <protocollo>
    <numero>1</numero>
    <data>2014-06-23</data>
    <oggetto/>
    <destinatario/>
    <operatore/>
  </protocollo>
     ...
</ProtocolloList>

我需要使用JAXB将此结果转换为Protocollo对象列表,以便我可以使用JList显示它们.因此,在这里的一个讨论后,我宣布了以下类:

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "protocollo")
public class Protocollo {

  private int numero;
  private String data;
  private String oggetto;
  private String destinatario;
  private String operatore;

  public Protocollo(String d, String o, String des, String op) {
    this.data = d;
    this.oggetto = o;
    this.destinatario = des;
    this.operatore = op;
  }

  public Protocollo() {

  }

  @XmlElement
  public int getNumero() {
      return numero;
  }

  public void setNumero(int numero) {
      this.numero = numero;
  }

  @XmlElement
  public String getData() {
      return data;
  }

  public void setData(String data) {
      this.data = data;
  }

  @XmlElement
  public String getOggetto() {
      return oggetto;
  }

  public void setOggetto(String oggetto) {
      this.oggetto = oggetto;
  }

  @XmlElement
  public String getDestinatario() {
      return destinatario;
  }

  public void setDestinatario(String destinatario) {
      this.destinatario = destinatario;
  }

  @XmlElement
  public String getOperatore() {
      return operatore;
  }

  public void setOperatore(String operatore) {
      this.operatore = operatore;
  }

}

和

import java.util.ArrayList;

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlElementWrapper;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "ProtocolloList")
public class ProtocolloList {

  @XmlElementWrapper(name = "ProtocolloList")
  @XmlElement(name = "protocollo")
  private ArrayList<Protocollo> ProtocolloList;

  public ArrayList<Protocollo> getProtocolloList() {
      return ProtocolloList;
  }

  public void setProtocolloList(ArrayList<Protocollo> protocolloList) {
      ProtocolloList = protocolloList;
  }
}

最后我执行了这样的转换:

JAXBContext jaxbContext = JAXBContext.newInstance(Protocollo.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
StringReader reader = new StringReader(this.resultXML);
protocolli = (ProtocolloList) unmarshaller.unmarshal(reader);

我继续得到这个例外:

unexpected element (uri:"", local:"ProtocolloList"). Expected elements are <{}protocollo>

我想我用注释犯了一些错误.你能帮我吗？

Answer 1

对于您的用例,您不需要@XmlElementWrapper注释.这是因为ProtocolList元素对应于您的@XmlRootElement注释.然后,您需要在@XmlElement属性上的注释来获取每个列表项.

@XmlRootElement(name = "ProtocolloList")
public class ProtocolloList {

  private ArrayList<Protocollo> ProtocolloList;

  @XmlElement(name = "protocollo")
  public ArrayList<Protocollo> getProtocolloList() {
      return ProtocolloList;
  }

}

注意:

默认情况下,您应该注释该属性.如果您想要注释您应该@XmlAccessorType(XmlAccessType.FIELD)在课堂上添加的字段.

UPDATE

您需要确保您JAXBContext了解根类.您可以将JAXBContext创建代码更改为以下内容:

JAXBContext jaxbContext = JAXBContext.newInstance(ProtocolloList.class);