Question

我遇到嵌套列表问题＆amp;我希望有人可以提供帮助。我想在下面的XML中解组列表。我希望有人可以指出我正确的方向，我相信我的注释是关闭的。当我尝试解组MyList数组时，它返回null。

我有一个XML：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<root>
    <myList>
        <name>N1</name>
        <type>T1</type>
        <version>V1</version>
    </myList>
    <myList>
        <name>N2</name>
        <type>T2</type>
        <version>V2</version>
    </myList>
</root>

MyList对象：

@XmlRootElement(name = "myList")
@XmlAccessorType(XmlAccessType.FIELD)
public class MyList {

  @XmlValue
  private String name;

  @XmlValue
  private String type;

  @XmlValue
  private String version;

Root Class：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement
public class Root {

  @XmlElementWrapper(name="myLists")
  @XmlElement(name="myList")
  private List<MyList> list = new ArrayList<MyList>();

非常感谢任何帮助。

Answer 1

@XmlValue错了。此外，如果@XmlElementWrapper元素未被包装，则myList错误。

试试这个，它对我有用。

<强> Root.java

package de.lhorn.so;

import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "root")
public class Root {

    @XmlElement(name = "myList")
    private List<MyList> list = new ArrayList<>();

    public Root() {
    }

    public List<MyList> getList() {
        return list;
    }

    public void setList(List<MyList> list) {
        this.list = list;
    }

    @Override
    public String toString() {
        return "Root{" + "list=" + list + '}';
    }

}

<强> MyList.java

package de.lhorn.so;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "myList")
@XmlAccessorType(XmlAccessType.FIELD)
public class MyList {

    private String name;

    private String type;

    private String version;

    public MyList() {
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getType() {
        return type;
    }

    public void setType(String type) {
        this.type = type;
    }

    public String getVersion() {
        return version;
    }

    public void setVersion(String version) {
        this.version = version;
    }

    @Override
    public String toString() {
        return "MyList{" + "name=" + name + ", type=" + type + ", version=" + version + '}';
    }

}

主要

InputStream is = SOPlayground.class.getResourceAsStream("root.xml"); JAXBContext jaxbContext = JAXBContext.newInstance(Root.class); Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller(); Root root = (Root) jaxbUnmarshaller.unmarshal(is); System.out.println(root);

<强>输出

Root{list=[MyList{name=N1, type=T1, version=V1}, MyList{name=N2, type=T2, version=V2}]}

Answer 2

试试这个，它应该有用。

@XmlRootElement(name = "root")
@XmlAccessorType (XmlAccessType.FIELD)
public class Root {

      @XmlElement(name="myList")
      private List<MyList> list = new ArrayList<MyList>();

}

<强>解组

 JAXBContext jaxbContext = JAXBContext.newInstance(Root.class);
        Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
        //We had written this file in marshalling example
        Root root= (Root) jaxbUnmarshaller.unmarshal( new File("c:/foo.xml") );
        List<MyList> myLists =   root.getList();

JAX-B嵌套列表

2 个答案: