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POEfficient way of parsing XML in Java
 

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  1. POEfficient way of parsing XML in Java
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    <p>I have to parse an XML file with following structure: </p> <pre><code>&lt;root&gt; &lt;object_1&gt; &lt;pro1&gt; abc &lt;/pro1&gt; &lt;pro2&gt; pqr &lt;/pro2&gt; &lt;pro3&gt; xyz &lt;/pro3&gt; &lt;children&gt; &lt;object_a&gt; &lt;pro1&gt; abc &lt;/pro1&gt; &lt;pro2&gt; pqr &lt;/pro2&gt; &lt;pro3&gt; xyz &lt;/pro3&gt; &lt;children&gt; . . . &lt;/children&gt; &lt;/object_a&gt; &lt;/children&gt; &lt;/object_1&gt; &lt;object_2&gt; . . . &lt;/object_n&gt; &lt;/root&gt; </code></pre> <p>Aim is to parse this multilevel nesting. A few classes are defined in Java.</p> <pre><code>Class Object_1 Class Object_2 . . . Class Object_N </code></pre> <p>with their respective properties. </p> <p>The following code is working for me, but then this is not the best way of doing things. </p> <pre><code>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&lt;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&lt;?&gt; obj=null; AbsModel model = null; for(int index=0;index&lt;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); } } </code></pre> <p>Is there a better way of parsing this XML. </p>
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    1. USchai
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    1. COThere's SAX or XPath!
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      1. USBuhake Sindi
    2. COYou can also use some XML deserialization (XStream, ...)
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    3. COI agree with the gentleman - but it kinda depends on your requirement of efficiency: XPath is probably not going to perform as much as SAX or your current solution. But it is probably less prone to errors and less work to maintain.
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