Question

尝试动态创建任何具有属性的元素现在能够创建没有属性的动态元素，需要帮助为动态创建的元素添加属性。

动态元素创建

  public class CustomElement {


        private  List<JAXBElement<String>> Elements;


        @XmlAnyElement
        public List<JAXBElement<String>> getElements() {
            return Elements;
        }

        public void setElements(List<JAXBElement<String>> elements) {
            Elements = elements;
        }


        public void setElements(Map<String, String>  myElements, String namespaceURI) {

               List<JAXBElement<String>> elements = new ArrayList<JAXBElement<String>>();
                for (Map.Entry<String, String> mapElement: myElements.entrySet()) 
                {               
                    JAXBElement jAXBElement=new JAXBElement(new QName(namespaceURI,mapElement.getKey()), 
                            String.class, mapElement.getValue());

                    elements.add(jAXBElement);
                }
            Elements = elements;
        }

//not working attr added to parent element not to current element
     private Map<QName, String> attr;

        @XmlAnyAttribute
        public Map<QName, String> getAttr() {
            return attr;
        }

        public void setAttr(Map<QName, String> attr) {
            this.attr = attr;
        }

    }

设定值

 Map<String, String> myElements =new HashMap<String,String>();  

             myElements.put("connectmrf ","");  
                setElements(myElements,"www.xxxxx.xxx/xxx/vmas");

             Map<QName, String> attr=new HashMap<QName,String>();   
                attr.put(new QName("Name"),"Amit");  
                attr.put(new QName("age"),"10");  

                setAttr(attr);

当前结果：

     <state  age="10" Name="Amit">
            <vmas:connectmrf ></vmas:connectmrf >
      </state>

预期结果：请建议

     <state>
            <vmas:connectmrf  age="10" Name="Amit" ></vmas:connectmrf >
      </state>

Answer 1

警告！：这只是一种快速而又肮脏的编组方式

您需要使用@XmlAnyAttribute创建包装类例如：

public class AnyXmlElement {
    @XmlAnyAttribute
    private Map<QName, String> attributes;
    @XmlAnyElement
    private List<Object> elements;

    public AnyXmlElement() {
        attributes = new LinkedHashMap<QName, String>();
        elements = new ArrayList<Object>();
    }

    public void addAttribute(QName name, String value) {
        attributes.put(name, value);
    }

    public void addElement(Object element) {
        elements.add(element);
    }
}

通过将AnyXmlElement包裹在JAXBElement

中来确保动态名称

public static JAXBElement<AnyXmlElement> toJAXBElement(QName qName, AnyXmlElement any) {
    return new JAXBElement<AnyXmlElement>(qName, AnyXmlElement.class, any);
}

为了测试，我们创建了一个简单的根类

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public static class Root {
    @XmlAnyElement
    public List<Object> any;
}

和程序

public static void main(String[] args) throws Exception {
    AnyXmlElement any = new AnyXmlElement();
    any.addAttribute(new QName("bar"), "hello");

    Root root = new Root();
    root.any = Arrays.asList(toJAXBElement(new QName("foo"), any));
    // It's important to add AnyXmlElement.class here
    JAXBContext jc = JAXBContext.newInstance(Root.class, AnyXmlElement.class);
    Marshaller m = jc.createMarshaller();
    m.marshal(root, System.out);
}

这是生成的xml：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?><root><foo bar="hello world"/></root>

如何在JAXB中为动态元素添加动态属性

1 个答案: