通过相同的节点比较Java中两个不同的XML文件

我有两个不同的XML文件要比较

假设我想比较这些XML文件的具体内容

例如：

每个XML文件都有一个公共节点，称为：

<BURAK>
   <burak1>
      <burak2>a<burak2>
<BURAK>
   <burak1>
      <burak2>c<burak2>

我找到了一个有用的代码来创建这个结构，但是这段代码给了我两个文件之间的所有差异。 我想要的是转到特定节点并比较该节点中的特定子节点。 我如何用Java编写这种代码

关于“svasa”的建议，我以这种方式更改了代码：

try
        {
            String firstValue = null;
            String secondValue = null;

            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            dbf.setNamespaceAware(true);
            DocumentBuilder db = dbf.newDocumentBuilder();

            Document doc = db.parse( new File( "/abc.xml" ) );

            XPathFactory xPathFactory = XPathFactory.newInstance();
            XPath xpath = xPathFactory.newXPath();

            XPathExpression expr = xpath.compile( "//TOC/STRUCTURE/TOC_NODE/NODE_NAME");

            Object exprValue = expr.evaluate( doc, XPathConstants.STRING );

            if ( exprValue != null )
            {
                firstValue = exprValue.toString();
            }

            Document doc1 = db.parse( new File( "/def.xml" ) );

            XPathFactory xPathFactory1 = XPathFactory.newInstance();
            XPath xpath1 = xPathFactory1.newXPath();

            XPathExpression expr1 = xpath1.compile( "//TOC/cac:STRUCTURE/cac:TOC_NODE/cac:NODE_NAME");

            Object exprValue1 = expr1.evaluate( doc1, XPathConstants.STRING );

            if ( exprValue1 != null )
            {
                secondValue = exprValue1.toString();


            }

            if ( firstValue != null && secondValue != null )
            {
                System.out.println( firstValue );
                System.out.println( secondValue );


            }
        }
        catch ( Exception e )
        {
            e.printStackTrace();
        }

     }
}

这不起作用，或者我无法让它起作用。如果这是我想要的正确结构。代码在我看来是合乎逻辑的。正如我在评论中提到的，我确实需要第二条路径。结构看起来很相似，但不完全相同。

使用。它很容易编码、编译xpath表达式并使用它从xml中提取所需的数据

代码为：

        try
        {
            String firstValue = null;
            String secondValue = null;

            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            dbf.setNamespaceAware(true);
            DocumentBuilder db = dbf.newDocumentBuilder();

            Document doc = db.parse( new File( "/abc.xml" ) );

            XPathFactory xPathFactory = XPathFactory.newInstance();
            XPath xpath = xPathFactory.newXPath();

            XPathExpression expr = xpath.compile( "//BURAK/burak1/burak2");

            Object exprValue = expr.evaluate( doc, XPathConstants.STRING );

            if ( exprValue != null )
            {
                firstValue = exprValue.toString();
            }

            doc = db.parse( new File( "/def.xml" ) );

            exprValue = expr.evaluate( doc, XPathConstants.STRING );

            if ( exprValue != null )
            {
                secondValue = exprValue.toString();
            }

            if ( firstValue != null && secondValue != null )
            {
                System.out.println( firstValue );
                System.out.println( secondValue );


            }
        }
        catch ( Exception e )
        {
            e.printStackTrace();
        }

xpath

//BURAK/burak1/burak2

将结果指定为

XPathConstants.STRING

因为您的两个XML相似，所以相同的xpath表达式可以工作

尝试执行代码，您将获得值

a

和

c

，然后您可以进行比较。

Try
 try
        {  File inputFile = new File("abc.xml");
         DocumentBuilderFactory dbFactory 
            = DocumentBuilderFactory.newInstance();
         DocumentBuilder dBuilder;

         dBuilder = dbFactory.newDocumentBuilder();

         Document doc = dBuilder.parse(inputFile);
         doc.getDocumentElement().normalize();


        XPath xPath =  XPathFactory.newInstance().newXPath();
     String expression="NODE";

     NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(doc, XPathConstants.NODESET);
     for (int i = 0; i < nodeList.getLength(); i++) {
           Node nNode = nodeList.item(i);
           System.out.println("\nCurrent Element :" 
                    + nNode.getNodeName());
          if (nNode.getNodeType() == Node.ELEMENT_NODE) {
              Element eElement = (Element) nNode;

              System.out.println("NODE_NAME: " 
                      + eElement
                         .getElementsByTagName("NODE")
                         .item(0)
                         .getTextContent());
              System.out.println("NODE: " 
                      + eElement
                         .getElementsByTagName("NODE")
                         .item(0)
                         .getTextContent());


              File inputFile1 = new File("def.xml");
             DocumentBuilderFactory dbFactory1 
                = DocumentBuilderFactory.newInstance();
             DocumentBuilder dBuilder1;

             dBuilder1 = dbFactory1.newDocumentBuilder();

             Document doc1 = dBuilder1.parse(inputFile1);
             doc1.getDocumentElement().normalize();


            XPath xPath1 =  XPathFactory.newInstance().newXPath();
          String expression1="NODE1";

          NodeList nodeList1 = (NodeList) xPath1.compile(expression1).evaluate(doc1, XPathConstants.NODESET);
          for (int j = 0; j < nodeList1.getLength(); j++) {
               Node nNode1 = nodeList1.item(j);
               System.out.println("\nCurrent Element :" 
                         + nNode1.getNodeName());
              if (nNode1.getNodeType() == Node.ELEMENT_NODE) {
                   Element eElement1 = (Element) nNode1;

                   System.out.println("NODE: " 
                           + eElement1
                              .getElementsByTagName("NODE_NAME")
                              .item(0)
                              .getTextContent());
                   System.out.println("NODE: " 
                           + eElement1
                              .getElementsByTagName("NODE_NAME")
                              .item(0)
                              .getTextContent());
        }

{File inputFile=新文件（“abc.xml”）；
DocumentBuilderFactory数据库工厂
=DocumentBuilderFactory.newInstance（）；
文档生成器dBuilder；
dBuilder=dbFactory.newDocumentBuilder（）；
Document doc=dBuilder.parse（inputFile）；
doc.getDocumentElement（）.normalize（）；
XPath=XPathFactory.newInstance（）.newXPath（）；
字符串表达式=“节点”；
NodeList NodeList=（NodeList）xPath.compile（expression.evaluate（doc，XPathConstants.NODESET）；
for（int i=0；i

---

这是我对这个问题的回答。感谢“svasa”的建议。

cod看起来很合理，但不幸的是它们并不完全相同，所以我应该为其他文件定义另一个xpath表达式来匹配它们。如果您使用第二个xml更新问题，我们可以查看是否需要另一个xpath

 try
        {  File inputFile = new File("abc.xml");
         DocumentBuilderFactory dbFactory 
            = DocumentBuilderFactory.newInstance();
         DocumentBuilder dBuilder;

         dBuilder = dbFactory.newDocumentBuilder();

         Document doc = dBuilder.parse(inputFile);
         doc.getDocumentElement().normalize();


        XPath xPath =  XPathFactory.newInstance().newXPath();
     String expression="NODE";

     NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(doc, XPathConstants.NODESET);
     for (int i = 0; i < nodeList.getLength(); i++) {
           Node nNode = nodeList.item(i);
           System.out.println("\nCurrent Element :" 
                    + nNode.getNodeName());
          if (nNode.getNodeType() == Node.ELEMENT_NODE) {
              Element eElement = (Element) nNode;

              System.out.println("NODE_NAME: " 
                      + eElement
                         .getElementsByTagName("NODE")
                         .item(0)
                         .getTextContent());
              System.out.println("NODE: " 
                      + eElement
                         .getElementsByTagName("NODE")
                         .item(0)
                         .getTextContent());


              File inputFile1 = new File("def.xml");
             DocumentBuilderFactory dbFactory1 
                = DocumentBuilderFactory.newInstance();
             DocumentBuilder dBuilder1;

             dBuilder1 = dbFactory1.newDocumentBuilder();

             Document doc1 = dBuilder1.parse(inputFile1);
             doc1.getDocumentElement().normalize();


            XPath xPath1 =  XPathFactory.newInstance().newXPath();
          String expression1="NODE1";

          NodeList nodeList1 = (NodeList) xPath1.compile(expression1).evaluate(doc1, XPathConstants.NODESET);
          for (int j = 0; j < nodeList1.getLength(); j++) {
               Node nNode1 = nodeList1.item(j);
               System.out.println("\nCurrent Element :" 
                         + nNode1.getNodeName());
              if (nNode1.getNodeType() == Node.ELEMENT_NODE) {
                   Element eElement1 = (Element) nNode1;

                   System.out.println("NODE: " 
                           + eElement1
                              .getElementsByTagName("NODE_NAME")
                              .item(0)
                              .getTextContent());
                   System.out.println("NODE: " 
                           + eElement1
                              .getElementsByTagName("NODE_NAME")
                              .item(0)
                              .getTextContent());
        }