读取xml的最快和最优化的方法

Question

我有以下XML文件

<book>
<person>
  <first>Kiran</first>
  <last>Pai</last>
  <age>22</age>
</person>
<person>
  <first>Bill</first>
  <last>Gates</last>
  <age>46</age>
</person>
<person>
  <first>Steve</first>
  <last>Jobs</last>
  <age>40</age>
</person>
</book>

现在,从XML文件读取数据的Java程序如下所示.

import java.io.File;
import org.w3c.dom.Document;
import org.w3c.dom.*;

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException; 

public class ReadAndPrintXMLFile{

    public static void main (String argv []){
    try {

            DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
            Document doc = docBuilder.parse (new File("book.xml"));

            // normalize text representation
            doc.getDocumentElement ().normalize ();
            System.out.println ("Root element of the doc is " + 
                 doc.getDocumentElement().getNodeName());


            NodeList listOfPersons = doc.getElementsByTagName("person");
            int totalPersons = listOfPersons.getLength();
            System.out.println("Total no of people : " + totalPersons);

            for(int s=0; s<listOfPersons.getLength() ; s++){


                Node firstPersonNode = listOfPersons.item(s);
                if(firstPersonNode.getNodeType() == Node.ELEMENT_NODE){


                    Element firstPersonElement = (Element)firstPersonNode;

                    //-------
                    NodeList firstNameList = firstPersonElement.getElementsByTagName("first");
                    Element firstNameElement = (Element)firstNameList.item(0);

                    NodeList textFNList = firstNameElement.getChildNodes();
                    System.out.println("First Name : " + 
                           ((Node)textFNList.item(0)).getNodeValue().trim());

                    //-------
                    NodeList lastNameList = firstPersonElement.getElementsByTagName("last");
                    Element lastNameElement = (Element)lastNameList.item(0);

                    NodeList textLNList = lastNameElement.getChildNodes();
                    System.out.println("Last Name : " + 
                           ((Node)textLNList.item(0)).getNodeValue().trim());

                    //----
                    NodeList ageList = firstPersonElement.getElementsByTagName("age");
                    Element ageElement = (Element)ageList.item(0);

                    NodeList textAgeList = ageElement.getChildNodes();
                    System.out.println("Age : " + 
                           ((Node)textAgeList.item(0)).getNodeValue().trim());

                    //------


                }//end of if clause


            }//end of for loop with s var


        }catch (SAXParseException err) {
        System.out.println ("** Parsing error" + ", line " 
             + err.getLineNumber () + ", uri " + err.getSystemId ());
        System.out.println(" " + err.getMessage ());

        }catch (SAXException e) {
        Exception x = e.getException ();
        ((x == null) ? e : x).printStackTrace ();

        }catch (Throwable t) {
        t.printStackTrace ();
        }
        //System.exit (0);

    }//end of main


}

结果是......

Root element of the doc is book
Total no of people : 3
First Name : Kiran
Last Name : Pai
Age : 22
First Name : Bill
Last Name : Gates
Age : 46
First Name : Steve
Last Name : Jobs
Age : 40

现在我的查询是请指教是否还有其他任何方式,这是最快的阅读这个xml,我看起来很好,请指教.. !!

Answer 1

使用ReadAndPrintXMLFileWithStAX下面的内容,当我与gontard给出ReadAndPrintXMLFileWithSAX的答案进行比较时,StAX方法更快.我的测试包括500000在JDK 1.7.0_07上为Mac 运行示例代码时间.

ReadAndPrintXMLFileWithStAX:  103 seconds
ReadAndPrintXMLFileWithSAX:   125 seconds

ReadAndPrintXMLFileWithStAX(使用Java SE 7)

下面是一个更优化的StAX(JSR-173)示例,XMLStreamReader而不是使用XMLEventReader.

import java.io.FileInputStream;
import java.io.InputStream;
import javax.xml.stream.*;

public class ReadAndPrintXMLFileWithStAX {

    public static void main(String argv[]) throws Exception {
        XMLInputFactory inputFactory = XMLInputFactory.newInstance();
        InputStream in = new FileInputStream("book.xml");
        XMLStreamReader streamReader = inputFactory.createXMLStreamReader(in);
        streamReader.nextTag(); // Advance to "book" element
        streamReader.nextTag(); // Advance to "person" element

        int persons = 0;
        while (streamReader.hasNext()) {
            if (streamReader.isStartElement()) {
                switch (streamReader.getLocalName()) {
                case "first": {
                    System.out.print("First Name : ");
                    System.out.println(streamReader.getElementText());
                    break;
                }
                case "last": {
                    System.out.print("Last Name : ");
                    System.out.println(streamReader.getElementText());
                    break;
                }
                case "age": {
                    System.out.print("Age : ");
                    System.out.println(streamReader.getElementText());
                    break;
                }
                case "person" : {
                    persons ++;
                }
                }
            }
            streamReader.next();
        }
        System.out.print(persons);
        System.out.println(" persons");
    }

}

产量

First Name : Kiran
Last Name : Pai
Age : 22
First Name : Bill
Last Name : Gates
Age : 46
First Name : Steve
Last Name : Jobs
Age : 40
3 persons

Answer 2

如果表演在您的情况下很重要,您应该更喜欢SAX或StAX(http://en.wikipedia.org/wiki/StAX)到DOM.

随着DOM,在第一时间内XML文件被解析成一个对象模型,那么你已经可以问它.所以对你来说算法有两遍.

随着SAX,在解析过程中,一些回调被调用(startDocument,endElement...), SAX是基于事件或推模式.

使用StAX,您可以控制解析.您将光标从元素移动到另一个元素.这是拉模型.

对于一个包含32910000人的文件,我将我的版本与SAX过度回答(Blaise Doughan)进行比较StAX.我删除所有的System.out.printlninstrusctions.我的程序花了106秒读取所有文件,另一个花了94秒.我认为这样做SAX比较慢,因为callback即使它们什么也不做(推模型)所有都被调用,而使用StAX时,光标只在"interresting"元素(拉模型)上移动.

例如,使用java 7:

import java.io.File;

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;

import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

public class ReadAndPrintXMLFileWithSax {

    public static void main(String[] args) throws Exception {
        SAXParserFactory fabrique = SAXParserFactory.newInstance();
        SAXParser parser = fabrique.newSAXParser();

        File file = new File("book.xml");
        BookHandler handler = new BookHandler();
        parser.parse(file, handler);
    }

    public static class BookHandler extends DefaultHandler {
        private int count = 0;
        private StringBuilder buffer;

        @Override
        public void startElement(String uri, String localName, String qName,
                Attributes attributes) throws SAXException {
            switch (qName) {
            case "person":
                count++;
                break;
            case "first":
                buffer = new StringBuilder("First Name : ");
                break;
            case "last":
                buffer = new StringBuilder("Last Name : ");
                break;
            case "age":
                buffer = new StringBuilder("Age : ");
                break;
            }
        }

        @Override
        public void characters(char[] ch, int start, int length)
                throws SAXException {
            String content = new String(ch, start, length);
            if (buffer != null)
                buffer.append(content);
        }

        @Override
        public void endElement(String uri, String localName, String qName)
                throws SAXException {
            switch (qName) {
            case "first":
            case "last":
            case "age":
                System.out.println(buffer.toString());
                break;
            }
        }

        @Override
        public void endDocument() throws SAXException {
            System.out.println(count + " persons");
        }
    }
}