Question

我正在通过StackOverflow寻找，但我找不到最佳答案。我需要在JAVA中读取XML文件。我的XML文件如下所示：`

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<server>
    <server01>
       <department>A1</department>
       <department>A2</department> 
    </server01>
    <server02>
       <department>A1</department>
       <department>A2</department> 
    </server02>
</server>

是否有可能在<server01>和<server02>之间读取Java元素，不包括文件的其余部分？或者我需要使用不同的标签？我找到了如何读取父节点的方法：

NodeList serversNames = xmlD.getDocumentElement().getChildNodes();

for (int i = 0; i < serversNames.getLength(); i++) {
    Node node = serversNames.item(i);
    if (node instanceof Element) {
        Element childElement = (Element) node;
        System.out.println("tag name: " + childElement.getTagName());
    }
}

我能够阅读所有部门标签并将其保存到阵列：

NodeList serverName = xmlD.getElementsByTagName("department");

serversList = new String[serverName.getLength()];

System.out.println("zasieg: " + serverName.getLength());

for (int temp = 0; temp < serverName.getLength(); temp++) {
    Element shareName = null;

    shareName = (Element) serverName.item(temp);

    serversList[temp] = shareName.getTextContent();
    System.out.println(temp + " - " + serversList[temp]);
}

那么，是否有可能只读取父节点的元素？只有SERVER01的元素？

Answer 1

XPath就是你想要的。对于所有意图和目的，您可以像sql一样考虑xpath，只考虑xml文档而不是数据库。这是一个使用Java的简单示例（请记住，xpath是一个标准，并不是特定于java，因此您可以在几乎任何流行语言中找到许多方法）：

    // Load document
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder = factory.newDocumentBuilder();
    Document doc = builder.parse( new FileInputStream( "/tmp/xml" ) );

    // Create XPath expression
    XPathFactory xPathfactory = XPathFactory.newInstance();
    XPath xpath = xPathfactory.newXPath();
    XPathExpression expr = xpath.compile( "//server01" );

    // Find node 'server01'
    Node node = (Node) expr.evaluate( doc, XPathConstants.NODE );
    if( node == null ) {
        System.out.println( "Node not found" );
        System.exit( 0 );
    }

    // Extract departments
    Element server01 = (Element) node;
    for( int k = 0 ; k < server01.getChildNodes().getLength() ; k++ ) {
        Node childNode = server01.getChildNodes().item( k );
        // Check if current node is a department node
        if( "department".equals( childNode.getNodeName() ) ) {
            System.out.println( childNode.getNodeName() + ": " + childNode.getTextContent().trim() );
        }
    }

是否有可能避免这种情况？

是的，只需将XPath表达式更改为仅接收所需的节点。

// Load document
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document doc = builder.parse( new FileInputStream( "/tmp/xml" ) );

// Create XPath expression
XPathFactory xPathfactory = XPathFactory.newInstance();
XPath xpath = xPathfactory.newXPath();
XPathExpression expr = xpath.compile( "//server01/department" );

// Find nodes 'department' under node 'server01'
NodeList node = (NodeList) expr.evaluate( doc, XPathConstants.NODESET );

// Extract departments
for( int k = 0 ; k < node.getLength() ; k++ ) {
    Node childNode = node.item( k );
    // Check if current node is a department node
    if( "department".equals( childNode.getNodeName() ) ) {
        System.out.println( "[" + k + "] " + childNode.getNodeName() + ": " + childNode.getTextContent().trim() );
    }
}

您必须收到下一个输出：

[0] department: A1
[1] department: A2

阅读XML元素的内部XML

1 个答案: