java - 如何使用java将一个XML文件拆分为多个XML文件

Question

我是第一次在 Java 中使用 XML 文件，我需要一些帮助。我正在尝试使用 Java 将一个 XML 文件拆分为多个 XML 文件

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<products>
    <product>
        <description>Sony 54.6" (Diag) Xbr Hx929 Internet Tv</description>
        <gtin>00027242816657</gtin>
        <price>2999.99</price>
        <orderId>2343</orderId>
        <supplier>Sony</supplier>
    </product>
    <product>
        <description>Apple iPad 2 with Wi-Fi 16GB - iOS 5 - Black
        </description>
        <gtin>00885909464517</gtin>
        <price>399.0</price>
        <orderId>2343</orderId>
        <supplier>Apple</supplier>
    </product>
    <product>
        <description>Sony NWZ-E464 8GB E Series Walkman Video MP3 Player Blue
        </description>
        <gtin>00027242831438</gtin>
        <price>91.99</price>
        <orderId>2343</orderId>
        <supplier>Sony</supplier>
    </product>
    <product>
        <description>Apple MacBook Air A 11.6" Mac OS X v10.7 Lion MacBook
        </description>
        <gtin>00885909464043</gtin>
        <price>1149.0</price>
        <orderId>2344</orderId>
        <supplier>Apple</supplier>
    </product>
    <product>
        <description>Panasonic TC-L47E50 47" Smart TV Viera E50 Series LED
            HDTV</description>
        <gtin>00885170076471</gtin>
        <price>999.99</price>
        <orderId>2344</orderId>
        <supplier>Panasonic</supplier>
    </product>
</products>

我正在尝试获取三个 XML 文档，例如:

 <?xml version="1.0" encoding="UTF-8"?>
<products>
        <product>
            <description>Sony 54.6" (Diag) Xbr Hx929 Internet Tv</description>
            <gtin>00027242816657</gtin>
            <price currency="USD">2999.99</price>
            <orderid>2343</orderid>
        </product>
        <product>
            <description>Sony NWZ-E464 8GB E Series Walkman Video MP3 Player Blue</description>
            <gtin>00027242831438</gtin>
            <price currency="USD">91.99</price>
            <orderid>2343</orderid>
        </product>
</products>

每个供应商一个。我怎样才能收到它？对此的任何帮助都会很棒。

Answer 1

确保将“inputFile”中的路径更改为您的文件以及输出部分:

StreamResult result = new StreamResult(new File("C:\xmls\" + supplier.trim() + ".xml"));

这是您的代码。

import java.io.File;
import java.util.ArrayList;
import java.util.List;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.OutputKeys;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

public class ExtractXml
{
    /**
     * @param args
     */
    public static void main(String[] args) throws Exception
    {
        String inputFile = "resources/products.xml";

        File xmlFile = new File(inputFile);
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xmlFile);

        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true); // never forget this!

        XPathFactory xfactory = XPathFactory.newInstance();
        XPath xpath = xfactory.newXPath();
        XPathExpression allProductsExpression = xpath.compile("//product/supplier/text()");
        NodeList productNodes = (NodeList) allProductsExpression.evaluate(doc, XPathConstants.NODESET);

        //Save all the products
        List<String> suppliers = new ArrayList<String>();
        for (int i=0; i<productNodes.getLength(); ++i)
        {
            Node productName = productNodes.item(i);

            System.out.println(productName.getTextContent());
            suppliers.add(productName.getTextContent());
        }

        //Now we create the split XMLs

        for (String supplier : suppliers)
        {
            String xpathQuery = "/products/product[supplier='" + supplier + "']";

            xpath = xfactory.newXPath();
            XPathExpression query = xpath.compile(xpathQuery);
            NodeList productNodesFiltered = (NodeList) query.evaluate(doc, XPathConstants.NODESET);

            System.out.println("Found " + productNodesFiltered.getLength() + 
                               " product(s) for supplier " + supplier);

            //We store the new XML file in supplierName.xml e.g. Sony.xml
            Document suppXml = dBuilder.newDocument();

            //we have to recreate the root node <products>
            Element root = suppXml.createElement("products"); 
            suppXml.appendChild(root);
            for (int i=0; i<productNodesFiltered.getLength(); ++i)
            {
                Node productNode = productNodesFiltered.item(i);

                //we append a product (cloned) to the new file
                Node clonedNode = productNode.cloneNode(true);
                suppXml.adoptNode(clonedNode); //We adopt the orphan :)
                root.appendChild(clonedNode);
            }

            //At the end, we save the file XML on disk
            TransformerFactory transformerFactory = TransformerFactory.newInstance();
            Transformer transformer = transformerFactory.newTransformer();
            transformer.setOutputProperty(OutputKeys.INDENT, "yes");
            DOMSource source = new DOMSource(suppXml);

            StreamResult result =  new StreamResult(new File("resources/" + supplier.trim() + ".xml"));
            transformer.transform(source, result);

            System.out.println("Done for " + supplier);
        }
    }

}