具有使用 JAXB 的属性和内容的 XML 元素
如何使用 JAXB 生成以下 XML?
<sport type="" gender="">
sport description
</sport>
如何使用 JAXB 生成以下 XML?
<sport type="" gender="">
sport description
</sport>
使用 注释类型和性别属性,并使用 描述属性:@XmlAttribute
@XmlValue
package org.example.sport;
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement
public class Sport {
@XmlAttribute
protected String type;
@XmlAttribute
protected String gender;
@XmlValue;
protected String description;
}
详细信息
正确的方案应该是:
<?xml version="1.0" encoding="UTF-8"?>
<schema xmlns="http://www.w3.org/2001/XMLSchema"
targetNamespace="http://www.example.org/Sport"
xmlns:tns="http://www.example.org/Sport"
elementFormDefault="qualified"
xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
jaxb:version="2.0">
<complexType name="sportType">
<simpleContent>
<extension base="string">
<attribute name="type" type="string" />
<attribute name="gender" type="string" />
</extension>
</simpleContent>
</complexType>
<element name="sports">
<complexType>
<sequence>
<element name="sport" minOccurs="0" maxOccurs="unbounded"
type="tns:sportType" />
</sequence>
</complexType>
</element>
为 SportType 生成的代码将为:
package org.example.sport;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlType;
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "sportType")
public class SportType {
@XmlValue
protected String value;
@XmlAttribute
protected String type;
@XmlAttribute
protected String gender;
public String getValue() {
return value;
}
public void setValue(String value) {
this.value = value;
}
public String getType() {
return type;
}
public void setType(String value) {
this.type = value;
}
public String getGender() {
return gender;
}
public void setGender(String value) {
this.gender = value;
}
}