我的XML看起来像这样:
<ConnProf ConnProfID="1111">
<ConnNum>1</ConnNum>
<IsMSPA>false</IsMSPA>
<IsArray>false</IsArray>
<IsDDOR>false</IsDDOR>
<Subsystem SSID="2222"ConnProfID="3333">
<SSName>AA</SSName>
<GenericSSName>AA</GenericSSName>
<ConnFuncAddr>aaa</ConnFuncAddr>
<DSSNum>22</DSSNum>
<isRemoved>false</isRemoved>
</Subsystem>
<Subsystem SSID="4444" ConnProfID="5555">
<SSName>BBBB</SSName>
<GenericSSName>BB</GenericSSName>
<ConnFuncAddr>bbbbbb</ConnFuncAddr>
<DSSNum>44</DSSNum>
<isRemoved>false</isRemoved>
</Subsystem>
我无法获得ConnNum,IsMSPA,IsArray和IsDDOR。 我试着用ConnNum:
//get ConnNum
Node n = doc.getFirstChild();
if (n.hasChildNodes())
System.out.println(n.getFirstChild().getNodeValue());
else
System.out.println(n.getNodeValue());
但是当我期待1时它只返回null。
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)
{
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
try
{
DocumentBuilder db = dbf.newDocumentBuilder();
for (int i = 1; i <= 8; i++)
{
Document doc = db.parse("file" + i + ".xml");
doc.getDocumentElement ().normalize ();
System.out.println ("Root element of the doc is " + doc.getDocumentElement().getNodeName());
//get ConnNum
Node n = doc.getFirstChild();
if (n.hasChildNodes())
System.out.println(n.getFirstChild().getNodeValue());
else
System.out.println(n.getNodeValue());
NodeList listOfSubsystems = doc.getElementsByTagName("Subsystem");
int totalSubsystems = listOfSubsystems.getLength();
if (totalSubsystems == 0)
continue;
else
{
System.out.println("Total number of subsystems : " + totalSubsystems + "\n");
Dish dish = new Dish();
for(int s=0; s < listOfSubsystems.getLength() ; s++)
{
Node firstPersonNode = listOfSubsystems.item(s);
if(firstPersonNode.getNodeType() == Node.ELEMENT_NODE)
{
Element firstPersonElement = (Element)firstPersonNode;
printElement(firstPersonElement, "SSName");
printElement(firstPersonElement, "GenericSSName");
printElement(firstPersonElement, "ConnFuncAddr");
printElement(firstPersonElement, "DSSNum");
printElement(firstPersonElement, "SCNum");
printElement(firstPersonElement, "SCAcronym");
printElement(firstPersonElement, "PassNum");
printElement(firstPersonElement, "FzCode");
printElement(firstPersonElement, "isRemoved");
System.out.println("------------------");
}
}
System.out.println("\n==============================");
}
}
}
catch(ParserConfigurationException pce)
{
pce.printStackTrace();
}
catch(SAXException se)
{
se.printStackTrace();
}
catch(IOException ioe)
{
ioe.printStackTrace();
}
}
public static void printElement(Element a, String name)
{
NodeList elementList = a.getElementsByTagName(name);
Element b = (Element)elementList.item(0);
if (b != null)
{
NodeList list = b.getChildNodes();
System.out.println( ((Node)list.item(0)).getNodeValue().trim() );
}
}
}
答案 0 :(得分:13)
也许第一个孩子不是你想象的那样。 Whitespace在XML中很重要,firstChild可能实际上是 text 节点。
Nodes有一个类型,你可以迭代所有检查 Element 节点类型的子节点来获取实际元素的句柄。
编辑:打印您所追求的值。它在 Element 节点上进行过滤,然后对每个节点的第一个子节点(包含文本的节点)进行过滤。
NodeList nodeList = n.getChildNodes();
for (int j = 0; j < nodeList.getLength(); j++) {
Node childNode = nodeList.item(j);
if (childNode.getNodeType() == Node.ELEMENT_NODE) {
System.out.println(childNode.getNodeName() + " " + childNode.getFirstChild().getNodeValue());
}
}
此外,正如@Steve Townsend正确写道,如果您使用的是Java 1.5或更高版本,则可以使用getTextContent()
而不是childNode.getFirstChild().getNodeValue()
。
答案 1 :(得分:5)
NodeList nodeList = n.getChildNodes();
for (int j = 0; j < nodeList.getLength(); j++) {
Node childNode = nodeList.item(j);
if (childNode instanceof Element) {
Element childElement = (Element) childNode;
System.out.println(childElement.getNodeName() + " " + childElement.getFirstChild().getNodeValue());
}
}
答案 2 :(得分:2)
您需要在此处使用getTextContent()。您可以先使用getNodeName()
(仅限调试版)以确保您处于正确的位置。
getNodeValue()
返回null
。有一个表here,用于描述每个可能上下文中getNode*
的结果。
答案 3 :(得分:0)
尝试拨打
doc.getDocumentElement()
而不是
doc.getFirstChild()
答案 4 :(得分:0)
如果我打算使用DOM解析器。我总是喜欢以下方法,因为我们不需要知道xml的每个标记并逐个打印它们。
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.OutputStreamWriter;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
public class RecDOMP {
public static void main(String[] args) throws Exception{
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setValidating(false);
DocumentBuilder db = dbf.newDocumentBuilder();
// replace following path with your input xml path
Document doc = db.parse(new FileInputStream(new File ("D:\\ambuj\\input.xml")));
// replace following path with your output xml path
File OutputDOM = new File("D:\\ambuj\\outapip1.txt");
FileOutputStream fostream = new FileOutputStream(OutputDOM);
OutputStreamWriter oswriter = new OutputStreamWriter (fostream);
BufferedWriter bwriter = new BufferedWriter(oswriter);
// if file doesnt exists, then create it
if (!OutputDOM.exists()) {
OutputDOM.createNewFile();}
visitRecursively(doc,bwriter);
bwriter.close(); oswriter.close(); fostream.close();
System.out.println("Done");
}
public static void visitRecursively(Node node, BufferedWriter bw) throws IOException{
// get all child nodes
NodeList list = node.getChildNodes();
for (int i=0; i<list.getLength(); i++) {
// get child node
Node childNode = list.item(i);
if (childNode.getNodeType() == Node.TEXT_NODE)
{
//System.out.println("Found Node: " + childNode.getNodeName()
// + " - with value: " + childNode.getNodeValue()+" Node type:"+childNode.getNodeType());
String nodeValue= childNode.getNodeValue();
nodeValue=nodeValue.replace("\n","").replaceAll("\\s","");
if (!nodeValue.isEmpty())
{
System.out.println(nodeValue);
bw.write(nodeValue);
bw.newLine();
}
}
visitRecursively(childNode,bw);
}
}
}