Java中解析XML的有效方法

Question

I have to parse an XML file with following structure:

<root>
    <object_1>
        <pro1> abc </pro1>
        <pro2> pqr </pro2>
        <pro3> xyz </pro3>

        <children>
            <object_a>
                <pro1> abc </pro1>
                <pro2> pqr </pro2>
                <pro3> xyz </pro3>

                <children>
                    .
                    .
                    .
                </children>
            </object_a>
        </children>     
    </object_1>
    <object_2> 
    .
    . 
    .
    </object_n>
</root>

Aim is to parse this multilevel nesting. A few classes are defined in Java.

Class Object_1
Class Object_2
.
.
.
Class Object_N

with their respective properties.

The following code is working for me, but then this is not the best way of doing things.

File file = new File(fileName);
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(file);
doc.getDocumentElement().normalize();

if(doc ==null) return;

Node node = doc.getFirstChild();

NodeList lst = node.getChildNodes();
Node children = null ; 

int len = lst.getLength();
for(int index=0;index<len;index++)
{
    Node child = lst.item(index);
    String name = child.getNodeName();
    if(name=="Name")
        name = child.getNodeValue();
    else if(name=="Comment")
        comment = child.getNodeValue());
    else if(name=="children")
        children = child;
    }

    if(children==null) return; 

    lst = children.getChildNodes();
    len = lst.getLength();
    Class<?> obj=null;
    AbsModel model = null;
    for(int index=0;index<len;index++)
    {
        Node childNode = lst.item(index);
        String modelName = childNode.getNodeName();
        try {
            obj = Class.forName(modelName);
        } catch (ClassNotFoundException e) {
            e.printStackTrace();
        }
        if(obj!=null)
            model = (AbsModel) obj.newInstance();
        else
            model = new GenericModel();

        model.restoreDefaultPropFromXML(childNode);
        addChild(model);
    }
}

Is there a better way of parsing this XML.

Answer 1

考虑使用JAXB ，它是Java自版本6以来的一部分。您应该能够几乎没有代码地将XML文件解析（“解组”）到您自己的类中，只需添加一些注释即可明确说明对象结构与您的对象之间的映射。 XML结构。

Answer 2

StAX and or JAXB is almost always the way to go.

If the XML is really dynamic (like attributes specify the property name) ie <prop name="property" value="" /> then you will need to use StAX only or live with what JAXB will map it to (a POJO with name and value properties) and post process.

Personally I find combining StAX and JAXB the best. I parse to the elements I want and then use JAXB to turn the element into a POJO.

See Also :

• My own utility library that will turn an XML Stream into an iterator of objects .
• Parsing very large XML files and marshalling to Java Objects
• http://tedone.typepad.com/blog/2011/06/unmarshalling-benchmark-in-java-jaxb-vs-stax-vs-woodstox.html

Answer 3

尽管JAXB可能是最佳选择，但我还要提到jOOX ，它提供了类似JQuery的API，并且使XML文档的使用非常愉快。