我想將基于 jaxb 的 xml 文件讀入我的面向對象結構?？梢哉f這是我的 xml 文件： <?xml version="1.0" encoding="utf-8" standalone="yes"?> <children xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <child xsi:type="girl"> <age>12</age> <isdancing>true</isdancing> </child> <child xsi:type="boy"> <age>10</age> <issoccerplayer>true</issoccerplayer> </child> </children>Children是某種包含多個子元素的包裝元素。孩子可以是 xsi : type 指定的男孩或女孩。這兩個類有一些共同的元素（如age）和一些不同（排除）的元素（如isdancing或issoccerplayer）要讀取文件，我有這個方法： public static void main( String[] args ) throws JAXBException { JAXBContext jaxbContext; jaxbContext = JAXBContext.newInstance(Children.class); Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller(); File file = new File("C:/test.xml"); if (!file.exists()) System.out.println("File does not exist"); Children children = (Children) jaxbUnmarshaller.unmarshal(file); System.out.println(children.toString()); }我的孩子們的班級是這樣的： @XmlRootElement(name="children") @XmlAccessorType(XmlAccessType.FIELD) public class Children { @XmlElement(name="child") private List<Child> childrenList; public List<Child> getChildren() { return childrenList; } public void setChildren(List<Child> children) {this.childrenList = children;} @Override public String toString() { return ReflectionToStringBuilder.toString(this); } }我的孩子班看起來像這樣： @XmlAccessorType(XmlAccessType.FIELD) public class Child { @XmlAttribute(name="xsi:type") private XsiType xsiType; private int age; @XmlElement(name = "isdancing") private boolean isDancing; } }我現在的問題是，輸出正常，但 Child-class 的元素 xsiType 始終為 null，否則最終會出現 IllegalAnnotationExceptions，這與 XmlTest.model.Child.xsiType 相關所以我預計設置任何類型的 @Xml-Annotation 都會出現錯誤。有人可以幫我找出錯誤嗎？目標是迭代孩子列表并在運行時（基于 xsiType）決定這是女孩還是男孩。
3 回答
千巷貓影
TA貢獻1829條經驗 獲得超7個贊
你不需要你的XsiType課。您可以直接使用String。
在你的Child類中xsiType，屬性應該如下所示。
@XmlAttribute(name?=?"type",?namespace?=?"? private?String?xsiType;
注意：在@XmlAttribute注釋中
使用
name = "type"（不帶前綴xsi:）指定
namespaceXML 中給定的參數xmlns:xsi="..."
順便說一句：?您最好使用常量，
而不是鍵入字符串。所以你的改進代碼會是這樣的："http://www.w3.org/2001/XMLSchema-instance"XMLConstants.W3C_XML_SCHEMA_INSTANCE_NS_URI
@XmlAttribute(name?=?"type",?namespace?=?XMLConstants.W3C_XML_SCHEMA_INSTANCE_NS_URI) private?String?xsiType;
慕無忌1623718
TA貢獻1744條經驗 獲得超4個贊
xsi 類型通常用于表達對具體類型的引用。Jaxb 可以使用 xsi 類型，無需進一步的解決方法。
創建一個Boy和 一個Girl擴展的類Children。（您可能需要使用 調整類型名稱@XmlType）。這樣，所有具有的元素都xsi:type=Girl將綁定到該類Girl
@XmlAccessorType(XmlAccessType.FIELD)
@XmlSeeAlso({ Boy.class, Girl.class }) // Either use @XmlSeeAlso to register classes in the JaxbContext
// or add them to the context directly
public class Child {
private int age;
@XmlElement(name = "isdancing")
private boolean isDancing;
@XmlElement(name = "issoccerplayer")
private boolean isSoccerPlayer;
// Getter and setter for all fields
}
@XmlType(name = "boy") // can be omitted if default value matches with the default value
public class Boy extends Child {
}
@XmlType(name = "girl")
public class Girl extends Child {
}
完整的獨立示例：
package jaxb;
import java.io.File;
import java.io.StringReader;
import java.util.List;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Unmarshaller;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlSeeAlso;
import javax.xml.bind.annotation.XmlType;
public class Inheritance {
public static void main(String[] args) throws JAXBException {
JAXBContext jaxbContext;
jaxbContext = JAXBContext.newInstance(Children.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
String x = "<?xml version=\"1.0\" encoding=\"utf-8\" standalone=\"yes\"?>\r\n"
+ " <children xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\">\r\n"
+ " <child xsi:type=\"girl\">\r\n" + " <age>12</age>\r\n"
+ " <isdancing>true</isdancing>\r\n" + " </child>\r\n"
+ " <child xsi:type=\"boy\">\r\n" + " <age>10</age>\r\n"
+ " <issoccerplayer>true</issoccerplayer>\r\n" + " </child>\r\n" + " </children>";
Children children = (Children) jaxbUnmarshaller.unmarshal(new StringReader(x));
System.out.println(children.getChildren().toString());
}
@XmlRootElement(name = "children")
@XmlAccessorType(XmlAccessType.FIELD)
public static class Children {
@XmlElement(name = "child")
private List<Child> childrenList;
public List<Child> getChildren() {
return childrenList;
}
public void setChildren(List<Child> children) {
this.childrenList = children;
}
}
@XmlAccessorType(XmlAccessType.FIELD)
@XmlSeeAlso({ Boy.class, Girl.class })
public static class Child {
private int age;
@XmlElement(name = "isdancing")
private boolean isDancing;
@XmlElement(name = "issoccerplayer")
private boolean isSoccerPlayer;
// Getter and setter for all fields
}
@XmlType(name = "boy")
public static class Boy extends Child {
}
@XmlType(name = "girl")
public static class Girl extends Child {
}
}
慕哥6287543
TA貢獻1831條經驗 獲得超10個贊
第二種方法的干凈解決方案（基于單獨的類文件）：
public class App
{
public static void main(String[] args) throws JAXBException
{
JAXBContext jaxbContext = JAXBContext.newInstance(Children.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
File file = new File("C:/test2.xml");
Children children = (Children) jaxbUnmarshaller.unmarshal(file);
for (Child c : children.getChildren()) {
if (c instanceof Boy) {
System.out.println(((Boy)c).toString());
} else if (c instanceof Girl){
System.out.println(((Girl)c).toString());
}
}
}
}
Children.java
@XmlRootElement(name="children")
@XmlAccessorType(XmlAccessType.FIELD)
public class Children {
@XmlElement(name="child")
private List<Child> childrenList;
public List<Child> getChildren() { return childrenList; }
public void setChildren(List<Child> children) {this.childrenList = children;}
@Override
public String toString() { return ReflectionToStringBuilder.toString(this); }
}
Boy.java
@XmlType(name="boy")
public class Boy extends Child {
@XmlElement(name = "issoccerplayer")
private boolean isSoccerPlayer;
public boolean isSoccerPlayer() { return isSoccerPlayer; }
public void setSoccerPlayer(boolean isSoccerPlayer) { this.isSoccerPlayer = isSoccerPlayer; }
@Override
public String toString() { return ReflectionToStringBuilder.toString(this); }
}
Girl.java
@XmlType(name="girl")
public class Girl extends Child {
@XmlElement(name = "isdancing")
private boolean isDancing;
public boolean isDancing() { return isDancing; }
public void setDancing(boolean isDancing) { this.isDancing = isDancing; }
@Override
public String toString() { return ReflectionToStringBuilder.toString(this); }
}
Child.java
@XmlAccessorType(XmlAccessType.FIELD)
@XmlSeeAlso({ Boy.class, Girl.class })
public abstract class Child {
private int age;
public int getAge() { return age; }
public void setAge(int age) { this.age = age; }
}
輸出應該是：
de.home.myproject.XmlTest.model.Girl@12edcd21[isDancing=true,age=12]
de.home.myproject.XmlTest.model.Boy@27bc2616[isSoccerPlayer=true,age=10]
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