Question

这是一项学术任务，我们获得了一个非常大的XML文件，其中包含数百个这样的条目。对于每个项目，我们应该列出经理的ID，将项目添加到列表的最后一个人的人员ID，以及当前的项目数。我已阅读并重读了Oracle DOM API和各种Node API。我们正在使用JAVA，我不能为我的生活找出如何搜索每个item_list节点的各个“字段”。以下是我们提供的数据示例。

<item_list id="item_list01">
   <numitems_intial>5</numitems_initial>
   <item>
     <date_added>1/1/2014</date_added>
     <added_by person="person01" />
   </item>
   <item>
      <date_added>1/6/2014</date_added>
      <added_by person="person05" />
    </item>
    <numitems_current>7</numitems_current>
    <manager person="person48" />
</item_list>
<item_list id="item_list02">
   <numitems_intial>5</numitems_initial>
   <item>
     <date_added>1/15/2014</date_added>
     <added_by person="person05" />
   </item>
   <item>
     <date_added>1/1/2014</date_added>
     <added_by person="person09" />
   </item>
       <item>
         <date_added>1/9/2014</date_added>
         <added_by person="person45" />
       </item>
        <numitems_current>7</numitems_current>
        <manager person="person38" />
    </item_list>

我尝试过类似的事情：

NodeList nodes = queryDoc.getElementsByTagName("item_list");

for(int i = 0; i < nodes.getLength(); i++) {

    Node node = nodes.item(i);

    if(node != null) {

       System.out.println(node.manager);

    }

}

暂时搞乱这段代码，但我想知道如何从每个节点的各个字段中检索数据。

Answer 1

如果您正在尝试阅读经理标签的人员属性，您可以按照以下所示进行操作 -

import java.io.IOException;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;


public class Test{

public static void main (String[] args)
{
    Test test = new Test();
    test.readXML();
}

private void readXML()
{
    Document doc = null;
    try 
    {
        doc = parseXML("/home/abc/Test.xml");
    } 
    catch (ParserConfigurationException e) 
    {
        e.printStackTrace();
    } 
    catch (SAXException e) 
    {
        e.printStackTrace();
    } 
    catch (IOException e) 
    {
        e.printStackTrace();
    }

    if(doc != null)
    {
        NodeList nList = doc.getElementsByTagName("item_list");
        for (int i = 0; i < nList.getLength(); i++) 
        {
            Node nNode = nList.item(i);
            Element eElement = (Element) nNode;
            Element cElement =  (Element) eElement.getElementsByTagName("manager").item(0);
            System.out.println("Manager ID : " + cElement.getAttribute("person"));
        }
    }
}

private Document parseXML(String filePath) throws ParserConfigurationException, SAXException, IOException
{
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    DocumentBuilder db = dbf.newDocumentBuilder();
    Document doc = db.parse(filePath);
    doc.getDocumentElement().normalize();
    return doc;
}
}

Answer 2

或者，使用xml，您可能需要编辑初始内容。我建议采用以下方法

import java.io.IOException;
import java.io.StringWriter;

import javax.xml.parsers.*;
import javax.xml.transform.*;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;

import org.w3c.dom.*;
import org.xml.sax.SAXException;


public class ReadXML {

    public static void main(String[] args) {
        try {
            Document doc = getDocument("/home/abc/Test.xml");

            System.out.println(getString(getNodeByName(doc,"item_list01")));
        } catch (TransformerException | ParserConfigurationException | IOException | SAXException e) {
            // Log e.printStackTrace();
        }
    }

    private static Document getDocument(String filePath) throws ParserConfigurationException, IOException, SAXException {
        DocumentBuilderFactory dbfac = DocumentBuilderFactory.newInstance();
        DocumentBuilder docBuilder = dbfac.newDocumentBuilder();

        return docBuilder.parse(filePath);
    }

    private static String getString(Node node) throws TransformerException {
        StringWriter sw = new StringWriter();
        Transformer t = TransformerFactory.newInstance().newTransformer();
        t.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
        t.setOutputProperty(OutputKeys.INDENT, "yes");
        t.transform(new DOMSource(node), new StreamResult(sw));

        return sw.toString();
    }

    public static Node getNodeByName(Document doc, String nodeName) {
        Node node = null;
        for (int i = 0; i < doc.getDocumentElement().getChildNodes().getLength(); i++) {
            if (!getTagName(doc, i).equals("#text")) {
                for (int j = 0; j < getNodeName(doc, i).getChildNodes().getLength(); j++) {
                    if (getNodeName(doc, i, j).equalsIgnoreCase("item_list")  && getNodeAttributes(doc,i,j).equalsIgnoreCase(nodeName)) {
                        node = getNodeName(doc, i);
                    }
                }
            }
        }
        return node;
    }

    private static String getTagName(Document doc, int i) {
        return getNodeName(doc, i).getNodeName();
    }

    private static Node getNodeName(Document doc, int i) {
        return (doc.getDocumentElement().getChildNodes().item(i));
    }

    private static String getNodeName(Document doc, int i, int j) {
        return getNodeName(doc, i).getChildNodes().item(j).getNodeName();
    }
    private static String getNodeAttributes(Document doc, int i, int j) {
        if(getNodeName(doc, i).getChildNodes().item(j).hasAttributes()){
            return getNodeName(doc, i).getChildNodes().item(j).getAttributes().item(0).getNodeValue();
        }
        return "";
    }
}

JAVA XML - 如何在XML节点中获取特定元素？

2 个答案: