如果有人帮助我,那就太好了。所以我有一个结构:
<letters>
<list>
<name>Simon</name>
<type>2</type>
<passengerName>Johny</passengerName>
<passengerSurname>Revelator</passengerSurname>
</list>
<list>
<name>someName</name>
<type>4</type>
<fileURL>someUrl</fileURL>
<specialNotes>specialNotes</specialNotes>
</list>
</letters>
对于这个结构,我编写了 Java 类,它们对它进行编组和解组:A 类包 com.edhex.testing;
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.PROPERTY)
@XmlSeeAlso({B.class, C.class})
@XmlType
abstract public class A {
int type;
String name;
@XmlElement
public int getType() {
return type;
}
public void setType(int type) {
this.type = type;
}
@XmlElement
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
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班级信件:
@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)
public class Letters {
List<A> list;
@XmlElement
public List<A> getList() {
return list;
}
public void setList(List<A> list) {
this.list = list;
}
}
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B类:
@XmlRootElement
@XmlType
@XmlAccessorType(XmlAccessType.PROPERTY)
public class B extends A {
String fileURL;
String specialNotes;
@XmlElement
public String getFileURL() {
return fileURL;
}
public void setFileURL(String fileURL) {
this.fileURL = fileURL;
}
@XmlElement
public String getSpecialNotes() {
return specialNotes;
}
public void setSpecialNotes(String specialNotes) {
this.specialNotes = specialNotes;
}
}
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C类:
@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)
public class C extends A {
String passengerName;
String passengerSurname;
@XmlElement
public String getPassengerName() {
return passengerName;
}
public void setPassengerName(String passengerName) {
this.passengerName = passengerName;
}
@XmlElement
public String getPassengerSurname() {
return passengerSurname;
}
public void setPassengerSurname(String passengerSurname) {
this.passengerSurname = passengerSurname;
}
}
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一切看起来都很好,但在编组之后我确实喜欢这样:
public static void main(String[] args) {
C c = new C();
c.setName("Simon");
c.setType(2);
c.setPassengerName("Johny");
c.setPassengerSurname("Revelator");
B b = new B();
b.setType(4);
b.setFileURL("someUrl");
b.setSpecialNotes("specialNotes");
b.setName("someName");
List<A> list = new ArrayList<A>(2);
list.add(c);
list.add(b);
Letters letter = new Letters();
letter.setList(list);
try {
File file = new File("C:\file.xml");
JAXBContext jaxbContext = JAXBContext.newInstance(Letters.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
jaxbMarshaller.marshal(letter, file);
jaxbMarshaller.marshal(letter, System.out);
} catch (JAXBException e) {
e.printStackTrace();
}
}
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我得到这样的输出:
<letters>
<list xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="c">
<name>Lasha</name>
<type>2</type>
<passengerName>Johny</passengerName>
<passengerSurname>Revelator</passengerSurname>
</list>
<list xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="b">
<name>someName</name>
<type>4</type>
<fileURL>someUrl</fileURL>
<specialNotes>specialNotes</specialNotes>
</list>
</letters>
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问题:如何删除每个列表标记中的 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="b" ?我知道 JAXB 会通知我它使用什么类型,因为最初映射使用另一种类型(A.class),但我不需要该信息。有人有什么想法吗?
@XmlElementRef如果您在属性上使用list,则该项目的元素将基于@XmlRootElement引用的类,然后您将无法获得属性xsi:type。
@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)
public class Letters {
List<A> list;
@XmlElementRef
public List<A> getList() {
return list;
}
public void setList(List<A> list) {
this.list = list;
}
}
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