Jaxb：如何解组xs：任何XML字符串部分？

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问题描述

我有一个应用程序使用Jaxb进行XML< - >转换，并使用maven-jaxb2-plugin自动生成类。

I have an application doing XML<->conversions using Jaxb and automatically generated classes with maven-jaxb2-plugin.

在我的架构深处，我有可能输入ANYxml。

Someplace deep in my schema, I have the possibility to enter "ANY" xml.

更新：这更好地描述了我的架构。一些已知的XML包含一个完全未知的部分（任何部分）。

Update: this better describes my schema. Some known XML wrapping a totally unknown part (the "any" part).

<xs:complexType name="MessageType">
  <xs:sequence>
    <xs:element name="XmlAnyPayload" minOccurs="0">
        <xs:complexType>
            <xs:sequence>
                <xs:any namespace="##any"/>
            </xs:sequence>
        </xs:complexType>
    </xs:element>
    <xs:element name="OtherElements">
        ....
</xs:sequence>

此地图（由jaxb提供）这样的内部类。

This maps (by jaxb) to a inner class like this.

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
    "any"
})
public static class XmlAnyPayload {

    @XmlAnyElement(lax = true)
    protected Object any;

当我解组整个结构时，没问题。 Object any将呈现为org.apache.xerces.dom.ElementNSImpl。现在，我想手动重新创建Java对象，然后转到XML。我如何获取一些随机XML并放入any（org.apache.xerces.dom.ElementNSImpl）元素以构建Java对象？

When I unmarshall the entire structure, it is no problem. The "Object any" will render into a org.apache.xerces.dom.ElementNSImpl. Now, I want to recreate the Java object manually and then go to XML. How do I take some random XML and put into the any (org.apache.xerces.dom.ElementNSImpl) element to be able to build up the Java object?

，接下来的情况是当我把这个元素作为java时，我想解组这个部分（能够提取这个元素的XML字符串）。但这是不可能的。我得到一个关于根元素的例外。但是无法注释ElementNSImpl。

Also, the next case is when I have this element as java, I want to unmarshall this very part (to be able to extract the XML string of this element). But this is not possible. I get an exception about root elements. But it is not possible to annotate ElementNSImpl.

unable to marshal type "com.sun.org.apache.xerces.internal.dom.ElementNSImpl" as an element because it is missing an @XmlRootElement annotation

你有什么建议吗？如何处理这些问题？

Do you have any suggestions on how to handle these problems?

推荐答案

@XmlAnyElement（lax = true）用简单的英语表示：

这正是发生的事情在你的情况下。因此，如果您想要实际解组此lax的内容，请为JAXB上下文提供您要解组的元素的映射。最简单的方法是使用 @XmlRootElement

This is exactly what is happening in your case. So if you want to actually unmarshal the content of this lax any, provide JAXB context with a mapping for the element you wish to unmarshal. The easiest way to do this is to annotate your class with @XmlRootElement

@XmlRootElement(name="foo", namespace="urn:bar")
public class MyClass { ... }

现在，当您创建JAXB上下文时，将 MyClass 添加到其中：

Now when you create your JAXB context, add MyClass into it:

JAXBContext context = JAXBContext.newInstance(A.class, B.class, ..., MyClass.class);

在这种情况下，如果JAXB符合 {urn：bar} foo 元素代替 xs：any ，它会知道这个元素被映射到 MyClass 并尝试解组MyClass。

In this case, if JAXB meets the {urn:bar}foo element in the place of that xs:any, it will know that this element is mapped onto MyClass and will try to unmarshal MyClass.

如果你是根据包名创建JAXB上下文（你可能会这样做），你仍然可以添加你的类（比方说， com.acme.foo.MyClass ）。最简单的方法是创建 com / acme / foo / jaxb.index 资源：

If you are creating JAXB context based on the package name (you probably do), you can still add you class (say, com.acme.foo.MyClass) to it. The easiest way is to create a com/acme/foo/jaxb.index resource:

com.acme.foo.MyClass

并将包名添加到上下文中路径：

And the add your package name to the context path:

JAXBContext context = JAXBContext.newInstance("org.dar.gee.schema:com.acme.foo");

还有其他方法可以使用 ObjectFactory 等。，但是 jaxb.index 的技巧可能是最简单的。

There are other ways with ObjectFactory etc., but the trick with jaxb.index is probably the easiest one.

或者，不要在一个中解组所有内容运行时，您可以将 xs：any 的内容保留为DOM，并在第二次解组时将其解组到目标对象中，并使用其他JAXB上下文（知道您的 MyClass 类）。类似于：

Alternatively, instead of unmarshalling everything in one run, you can leave the content of xs:any as DOM and unmarshal it into the target object in a second unmarshalling with anothe JAXB context (which know your MyClass class). Something like:

JAXBContext payloadContext = JAXBContext.newInstance(MyClass.class);
payloadContext.createUnmarshaller().unmarshal((Node) myPayload.getAny());

这种方法有时更好，特别是当你有相对独立的容器/有效载荷模式的组合时。取决于案例。

This approach is sometimes better, especially when you have a combination of container/payload schemas which are relatively independent. Depends on the case.

上述所有内容也适用于编组。它完全是双向的。

All said above applies to marshalling as well. It's all neatly bidirectional.

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