Java XML获取具有相同名称的所有节点

我的xml文档如下所示：

<?xml version="1.0"?>
<root>
    <success>true</success>
    <note>
        <note_id>32219</note_id>
        <the_date>1336763490</the_date>
        <member_id>108649</member_id>
        <area>6</area>
        <note>Note 123123123</note>
    </note>
    <note>
        <note_id>33734</note_id>
        <the_date>1339003652</the_date>
        <member_id>108649</member_id>
        <area>1</area>
        <note>This is another note.</note>
    </note>
    <note>
        <note_id>49617</note_id>
        <the_date>1343050791</the_date>
        <member_id>108649</member_id>
        <area>1</area>
        <note>this is a 3rd note.</note>
    </note>
</root>

真的
32219
1336763490
108649
6.
附注123123
33734
1339003652
108649
1.
这是另一个音符。
49617
1343050791
108649
1.
这是第三个音符。

我希望获取该文档，获取所有

标记并将它们转换为字符串，然后将它们传递给我的XML类，并将XML类放入数组列表中。我希望这是有道理的。下面是我试图用来获取所有

标记的方法

public ArrayList<XML> getNodes(String root, String name){
    ArrayList<XML> elList = new ArrayList<>();
    NodeList nodes = doc.getElementsByTagName(root);
    for(int i = 0; i < nodes.getLength(); i++){
        Element element = (Element)nodes.item(i);
        NodeList nl = element.getElementsByTagName(name);
        for(int c = 0; c < nl.getLength(); c++){
            Element e = (Element)nl.item(c);
            String xmlStr = this.nodeToString(e);
            XML xml = new XML();
            xml.parse(xmlStr);
            elList.add(xml);
        }
    }
    return elList;
}

private String nodeToString(Node node){
    StringWriter sw = new StringWriter();
    try{
        Transformer t = TransformerFactory.newInstance().newTransformer();
        t.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
        t.transform(new DOMSource(node), new StreamResult(sw));
    }catch(TransformerException te){
        System.out.println("nodeToString Transformer Exception");
    }
    return sw.toString();
}

public ArrayList getNodes（字符串根，字符串名）{
ArrayList elList=新的ArrayList（）；
NodeList节点=doc.getElementsByTagName（根）；
对于（int i=0；i

所以，我的问题是，如何将每个

标记作为字符串？有了这些代码，我现在得到的是

null

for

String xmlStr=e.getNodeValue（）

编辑

我编辑了我的主代码，这似乎有效
 使用XPath，以下代码：

public class NotesExtractor {
   public static List< String > getTextOf( Document doc, String tagName )
      throws Exception
   {
      List< String > notes  = new ArrayList<>();
      XPathFactory xPathFactory = XPathFactory.newInstance();
      XPath xPath = xPathFactory.newXPath();
      NodeList xText =
         (NodeList)xPath.evaluate(
            "//" + tagName + "/text()", doc, XPathConstants.NODESET );
      for( int i = 0; i < xText.getLength(); ++i ) {
         Text textElt = (Text)xText.item( i );
         String noteTxt = textElt.getTextContent().trim();
         if( ! noteTxt.isEmpty())
         {
            notes.add( noteTxt.trim());
         }
      }
      return notes;
   }

   public static void main( String[] args ) throws Exception {
      DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
      dbf.setIgnoringElementContentWhitespace( true );
      DocumentBuilder db = dbf.newDocumentBuilder();
      Document doc = db.parse( "Notes.xml" );
      System.out.println( getTextOf( doc, "note" ));
   }

澄清后更新

您可以使用XPath查找所有
元素

这将允许您简单地隔离每个节点。然后，您可以基于找到的节点创建一个新文档，并将其转换回字符串

public class TestXML01 {

    public static void main(String[] args) {

        String xml = "<?xml version=\"1.0\"?>";
        xml += "<root>";
        xml += "<success>true</success>";
        xml += "<note>";
        xml += "<note_id>32219</note_id>";
        xml += "<the_date>1336763490</the_date>";
        xml += "<member_id>108649</member_id>";
        xml += "<area>6</area>";
        xml += "<note>Note 123123123</note>";
        xml += "</note>";
        xml += "<note>";
        xml += "<note_id>33734</note_id>";
        xml += "<the_date>1339003652</the_date>";
        xml += "<member_id>108649</member_id>";
        xml += "<area>1</area>";
        xml += "<note>This is another note.</note>";
        xml += "</note>";
        xml += "<note>";
        xml += "<note_id>49617</note_id>";
        xml += "<the_date>1343050791</the_date>";
        xml += "<member_id>108649</member_id>";
        xml += "<area>1</area>";
        xml += "<note>this is a 3rd note.</note>";
        xml += "</note>";
        xml += "</root>";

        ByteArrayInputStream bais = null;

        try {
            bais = new ByteArrayInputStream(xml.getBytes());
            DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
            factory.setNamespaceAware(false);
            DocumentBuilder builder = factory.newDocumentBuilder();
            Document xmlDoc = builder.parse(bais);

            Node root = xmlDoc.getDocumentElement();

            XPathFactory xFactory = XPathFactory.newInstance();
            XPath xPath = xFactory.newXPath();

            XPathExpression xExpress = xPath.compile("/root/note");
            NodeList nodes = (NodeList) xExpress.evaluate(root, XPathConstants.NODESET);

            System.out.println("Found " + nodes.getLength() + " note nodes");

            for (int index = 0; index < nodes.getLength(); index++) {
                Node node = nodes.item(index);
                Document childDoc = builder.newDocument();
                childDoc.adoptNode(node);
                childDoc.appendChild(node);
                System.out.println(toString(childDoc));
            }

        } catch (Exception exp) {
            exp.printStackTrace();
        } finally {
            try {
                bais.close();
            } catch (Exception e) {
            }
        }
    }

    public static String toString(Document doc) {

        String sValue = null;

        ByteArrayOutputStream baos = null;
        OutputStreamWriter osw = null;

        try {
            baos = new ByteArrayOutputStream();
            osw = new OutputStreamWriter(baos);

            Transformer tf = TransformerFactory.newInstance().newTransformer();
            tf.setOutputProperty(OutputKeys.INDENT, "yes");
            tf.setOutputProperty(OutputKeys.METHOD, "xml");
            tf.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");

            DOMSource domSource = new DOMSource(doc);
            StreamResult sr = new StreamResult(osw);
            tf.transform(domSource, sr);

            osw.flush();
            baos.flush();
            sValue = new String(baos.toByteArray());
        } catch (Exception exp) {
            exp.printStackTrace();
        } finally {
            try {
                osw.close();
            } catch (Exception exp) {
            }
            try {
                baos.close();
            } catch (Exception exp) {
            }
        }
        return sValue;
    }
}

公共类TestXML01{
公共静态void main（字符串[]args）{
字符串xml=”“；
xml+=“”；
xml+=“true”；
xml+=“”；
xml+=“32219”；
xml+=“1336763490”；
xml+=“108649”；
xml+=“6”；
xml+=“注释123123”；
xml+=“”；
xml+=“”；
xml+=“33734”；
xml+=“1339003652”；
xml+=“108649”；
xml+=“1”；
xml+=“这是另一个注释。”；
xml+=“”；
xml+=“”；
xml+=“49617”；
xml+=“1343050791”；
xml+=“108649”；
xml+=“1”；
xml+=“这是第三个注释。”；
xml+=“”；
xml+=“”；
ByteArrayInputStream bais=null；
试一试{
bais=newbytearrayinputstream（xml.getBytes（））；
DocumentBuilderFactory工厂=DocumentBuilderFactory.newInstance（）；
factory.setNamespaceAware（false）；
DocumentBuilder=factory.newDocumentBuilder（）；
文档xmlDoc=builder.parse（BAI）；
节点根=xmlDoc.getDocumentElement（）；
XPathFactory xFactory=XPathFactory.newInstance（）；
XPath=xFactory.newXPath（）；
xPath表达式xExpress=xPath.compile（“/root/note”）；
NodeList节点=（NodeList）xExpress.evaluate（根，XPathConstants.NODESET）；
System.out.println（“找到”+节点.getLength（）+注释节点”）；
对于（int index=0；index

现在输出

Found 3 note nodes
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>32219</note_id>
    <the_date>1336763490</the_date>
    <member_id>108649</member_id>
    <area>6</area>
    <note>Note 123123123</note>
</note>

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>33734</note_id>
    <the_date>1339003652</the_date>
    <member_id>108649</member_id>
    <area>1</area>
    <note>This is another note.</note>
</note>

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>49617</note_id>
    <the_date>1343050791</the_date>
    <member_id>108649</member_id>
    <area>1</area>
    <note>this is a 3rd note.</note>
</note>

找到3个注释节点
32219
1336763490
108649
6.
附注123123
33734
1339003652
108649
1.
这是另一个音符。
49617
1343050791
108649
1.
这是第三个音符。
um。。嘿，问题是什么？@RohitJain更新了问题。请参阅底部。那么，您已经有了一个已解析的dom树，您不想用从该dom树检索到的值填充XML对象，而是想将该dom树的一个分支转换回一个字符串，然后重新解析该字符串？对每个音符元素都这样做？
Found 3 note nodes
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>32219</note_id>
    <the_date>1336763490</the_date>
    <member_id>108649</member_id>
    <area>6</area>
    <note>Note 123123123</note>
</note>

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>33734</note_id>
    <the_date>1339003652</the_date>
    <member_id>108649</member_id>
    <area>1</area>
    <note>This is another note.</note>
</note>

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<note>
    <note_id>49617</note_id>
    <the_date>1343050791</the_date>
    <member_id>108649</member_id>
    <area>1</area>
    <note>this is a 3rd note.</note>
</note>