Java - JAXB - Unmarshalling：如何实现正确的设置？

Question

我目前正在尝试将大量 XML 更改为 Java 对象，但我一直卡住。我尝试在网上复制很多不同的示例，但我似乎永远无法找到正确的方法，而且我发现很难调试。

我的 XML 看起来像这样

<?xml version="1.0" encoding="utf-8"?>
<XMLSOCCER.COM>
  <TeamLeagueStanding xmlns="http://xmlsoccer.com/LeagueStanding">
    <Team>Leicester</Team>
    <Team_Id>31</Team_Id>
    <Played>26</Played>
    <PlayedAtHome>12</PlayedAtHome>
    <PlayedAway>14</PlayedAway>
    <Won>15</Won>
    <Draw>8</Draw>
    <Lost>3</Lost>
    <NumberOfShots>464</NumberOfShots>
    <YellowCards>40</YellowCards>
    <RedCards>1</RedCards>
    <Goals_For>48</Goals_For>
    <Goals_Against>29</Goals_Against>
    <Goal_Difference>19</Goal_Difference>
    <Points>53</Points>
  </TeamLeagueStanding>
  <TeamLeagueStanding xmlns="http://xmlsoccer.com/LeagueStanding">
    <Team>Tottenham</Team>
    <Team_Id>21</Team_Id>
...

所以我只有一个 TeamLeagueStanding 列表，我想将其保留为 Team 对象。我的 Team 类的 java 代码目前是这样的

@XmlRootElement(name = "TeamLeagueStanding")
public class Team {

    @XmlElement(name = "Team")
    String teamName;
    @XmlElement(name = "Team_Id")
    int teamID;

    public Team (String team, int id) {
        super();
        this.teamName = team;
        this.teamID = id;
    }

}

我的 Teams 类就是用来保存 Teams 列表的

@XmlRootElement(name = "XMLSOCCER.COM")
public class Teams {

    @XmlElement
    List<Team> teamList;

    public Teams () {

    }

}

而我的主要功能是这样的

public class Main {

    public static void main(String[] args) throws Exception {

        File xml = new File("data/GetLeagueStandingsPrem1516.xml");
        JAXBContext jc = JAXBContext.newInstance(Teams.class);
        Unmarshaller um = jc.createUnmarshaller();

        Teams t = (Teams) um.unmarshal(xml);

        System.out.println(t.teamList.size());

    }

}

我已经尝试了很多方法，但总是得到空指针异常或各种IllegalAnnotationExceptions。如果有人知道我哪里出错了，我将不胜感激！

谢谢，

西蒙

Answer 1

我认为你应该添加@XmlAccessorType(XmlAccessType.FIELD) 注释

 @XmlRootElement(name = "TeamLeagueStanding")
 @XmlAccessorType(XmlAccessType.FIELD)
 public class Team {

  }

Answer 2

您的根元素似乎是 XMLSOCCER.COM
1。创建 xsd
2.使用xjc命令创建java bean类或者使用IDE创建对象。
3.keep在src下创建的xml bean对象
4.然后你传递要解组的xml
5. 您将获得可以访问的分层转换对象。

Answer 3

1. 为Team类添加默认构造函数

2. 添加

   @XmlElement(name = "TeamLeagueStanding")
   List<Team> teamList;
   

   到Teams类；

---

Team.class：

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

@XmlRootElement(name = "TeamLeagueStanding")
public class Team {

    @XmlElement(name = "Team")
    String teamName;
    @XmlElement(name = "Team_Id")
    int teamID;

    public Team (){

    }

    public Team (String team, int id) {
        super();
        this.teamName = team;
        this.teamID = id;
    }
}

Teams.class

import javax.xml.bind.annotation.*;
import java.util.List;

@XmlRootElement(name = "XMLSOCCER.COM")
@XmlAccessorType(XmlAccessType.FIELD)
public class Teams {

    @XmlElement(name = "TeamLeagueStanding")
    List<Team> teamList;

    public Teams () {

    }

}