2017-04-17 68 views
0

我有3个输入XML,它们几乎都具有相同的元素和属性,实际上它们表示相同的东西,所以我想将它们编组到相同的对象,类似于这样的:JAXB:解组不同的xml到同一个对象

请求一:

<?xml version="1.0" encoding="UTF-8"?> 
<RequestOne> 
    <id>123</id> 
    <name>foo</name> 
</RequestOne> 

请求二:

<?xml version="1.0" encoding="UTF-8"?> 
<RequestTwo> 
    <id>123</id> 
    <value>val</value> 
</RequestTwo> 

要求三:

<?xml version="1.0" encoding="UTF-8"?> 
<RequestThree> 
    <name>foo</name> 
    <value>val</value> 
</RequestThree> 

所需的对象(像):

@XmlRootElement 
public class Resource{ 

    @XmlElement 
    private String id; 
    @XmlElement 
    private String name; 
    @XmlElement 
    private String value; 

    //(...) more code 
} 

但我不能使用多个rootElement的注解问JAXB所有3请求解组的类Resource

的对象有一种方法来做到这一点?或者我必须做3个Sepparated类?

感谢您使用重载的通用unmarshal方法的帮助

回答

0

选项1

解组:

public static class Base { 
    private String name ; 

    @XmlElement(name = "name") 
    public String getName() { 
     return name; 
    } 

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

public static void main (String [] args) throws JAXBException { 
    JAXBContext jaxbContext = JAXBContext.newInstance(Base.class); 
    Unmarshaller unmarshaller = jaxbContext.createUnmarshaller(); 
    JAXBElement<Base> basea = unmarshaller.unmarshal(new StreamSource(new StringReader("<RootA><name>nanana</name></RootA>")), Base.class); 
    System.out.println(basea.getValue().getName()); 
    JAXBElement<Base> baseb = unmarshaller.unmarshal(new StreamSource(new StringReader("<RootB><name>nbnbnb</name></RootB>")), Base.class); 
    System.out.println(baseb.getValue().getName()); 
} 

选项2

您可以随时使用Java的类子类型功能吗? JAXB也对父类进行注释扫描。这个例子的作品

public static class Base { 
    private String name ; 

    @XmlElement(name = "name") 
    public String getName() { 
     return name; 
    } 

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

@XmlRootElement(name = "RootA") 
public static class RootA extends Base{ 
} 

@XmlRootElement(name = "RootB") 
public static class RootB extends Base { 
} 


public static void main (String [] args) throws JAXBException { 
    JAXBContext jaxbContext = JAXBContext.newInstance(RootA.class,RootB.class); 
    Unmarshaller unmarshaller = jaxbContext.createUnmarshaller(); 
    RootA rootA = (RootA)unmarshaller.unmarshal(new StringReader("<RootA><name>nanana</name></RootA>")); 
    System.out.println(rootA.getName()); 
    RootB rootB = (RootB)unmarshaller.unmarshal(new StringReader("<RootB><name>nbnbnb</name></RootB>")); 
    System.out.println(rootB.getName()); 
} 
+0

非常感谢,我用第一个选项,并完美地工作:D – Numpakov

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