我有一个像这样的xml:
<todo>
<doLaundry cost="1"/>
<washCar cost="10"/>
<tidyBedroom cost="0" experiencePoints="5000"/>
</todo>
它的XSD架构是:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:complexType name="todo">
<xs:sequence>
<xs:choice maxOccurs="unbounded">
<xs:element name="doLaundry" type="doLaundry" />
<xs:element name="washCar" type="washCar" />
<xs:element name="tidyBedroom" type="tidyBedroom" />
</xs:choice>
</xs:sequence>
</xs:complexType>
<xs:complexType name="doLaundry">
<xs:attribute name="cost" type="xs:int" />
</xs:complexType>
<xs:complexType name="washCar">
<xs:attribute name="cost" type="xs:int" />
</xs:complexType>
<xs:complexType name="tidyBedroom">
<xs:attribute name="cost" type="xs:int" />
<xs:attribute name="experiencePoints" type="xs:int" />
</xs:complexType>
</xs:schema>
当我通过JAXB处理这个模式时,我得到一个类如下的方法:
public class Todo {
public List<Object> getDoLaundryOrWashCarOrTidyBedroom() {
...
}
}
理想情况下,我想要的是一种定义所有其他XSD类型扩展的通用基类型的方法。从XSD架构生成的Jaxb类应该有一个返回通用任务列表的方法。这样可以很容易地将新任务添加到待办事项列表中:
public class Todo {
public List<Task> getTasks() {
...
}
}
public abstract class Task {
public int getCost() {
...
}
}
public class TidyBedroom extends Task {
public int getExperiencePoints() {
...
}
}
为了生成上述Java类,XSD架构应该是什么样的?
答案 0 :(得分:19)
我在Blaise Doughan的文章的帮助下找到了答案:http://bdoughan.blogspot.com/2010/11/jaxb-and-inheritance-using-xsitype.html
此架构:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:complexType name="todo">
<xs:sequence>
<xs:choice maxOccurs="unbounded">
<xs:element name="doLaundry" type="doLaundry" />
<xs:element name="washCar" type="washCar" />
<xs:element name="tidyBedroom" type="tidyBedroom" />
</xs:choice>
</xs:sequence>
</xs:complexType>
<xs:complexType abstract="true" name="Task">
<xs:attribute name="cost" type="xs:int" use="required" />
</xs:complexType>
<xs:complexType name="doLaundry">
<xs:complexContent>
<xs:extension base="Task">
</xs:extension>
</xs:complexContent>
</xs:complexType>
<xs:complexType name="washCar">
<xs:complexContent>
<xs:extension base="Task">
</xs:extension>
</xs:complexContent>
</xs:complexType>
<xs:complexType name="tidyBedroom">
<xs:complexContent>
<xs:extension base="Task">
<xs:attribute name="experiencePoints" type="xs:int" />
</xs:extension>
</xs:complexContent>
</xs:complexType>
</xs:schema>
结合绑定文件:
<jxb:bindings version="1.0" xmlns:jxb="http://java.sun.com/xml/ns/jaxb" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<jxb:bindings>
<jxb:bindings schemaLocation="todo.xsd" node="/xs:schema/xs:complexType[@name='todo']/xs:sequence/xs:choice">
<jxb:property name="Tasks"/>
</jxb:bindings>
</jxb:bindings>
</jxb:bindings>
将提供我在问题中描述的抽象和继承类。绑定文件将Jaxb的默认方法名称从getDoLaundryOrWashCarOrTidyBedroom()更改为getTasks()。
答案 1 :(得分:6)
xsd:choice对应于@XmlElements注释。您可以将此注释直接应用于所需的对象模型。
有关详细信息,请参阅:
答案 2 :(得分:0)
在架构中使用xs:extension,您将在架构中定义继承(扩展)JAXB类。
答案 3 :(得分:0)
也许我没有'得到'这个问题,但是有什么问题......
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:complexType name="todo">
<xs:sequence>
<xs:choice maxOccurs="unbounded">
<xs:element name="doLaundry" type="task" />
<xs:element name="washCar" type="task" />
<xs:element name="tidyBedroom" type="task" />
</xs:choice>
</xs:sequence>
</xs:complexType>
<xs:complexType name="task">
<xs:attribute name="cost" type="xs:int" />
<xs:attribute name="experiencePoints" type="xs:int" />
</xs:complexType>
</xs:schema>