将新的 XML 节点保存到文件

Question

我正在尝试将包含XML 的nodeList 节点保存为新文件，这是获取新XML 文档并拆分为更小的XMLs 的节点列表：

public void split(Document inDocument) throws ParserConfigurationException,
            SAXException, IOException {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = factory.newDocumentBuilder();
        SaveXML savePerJob = new SaveXML();
        // Load the input XML document, parse it and return an instance of the
        // Document class.
        Document document = inDocument;
        //all elements
        //jobs
        NodeList nodes = document.getDocumentElement().getChildNodes();
        NodeList jobs = nodes.item(7).getChildNodes();
        for(int j =0; j<jobs.getLength(); j++){
            Node itm = jobs.item(j);
            String itmName = itm.getFirstChild().getNodeName();
            String itmID = itm.getFirstChild().getTextContent();
            System.out.println("name: " + itmName + " value: " + itmID);
            //here I want to save the node as a new .xml file
        }
    }

输出是一个很长的列表，例如：

name: id value: 9496425

现在我想将节点 itm 保存为新的 .xml 文件，但我没有找到按原样返回节点的函数。 谢谢。

Answer 1

您可以将节点转换为字符串并将此字符串保存到.xml 文件中。

将节点转换为字符串

下面的方法会将一个节点变成一个xml字符串。这是一个 JDK 唯一的解决方案，不需要依赖项。

import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

import javax.xml.transform.OutputKeys;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
import java.io.StringWriter;

public static String toString(Node node, boolean omitXmlDeclaration, boolean prettyPrint) {
    if (node == null) {
        throw new IllegalArgumentException("node is null.");
    }

    try {
        // Remove unwanted whitespaces
        node.normalize();
        XPath xpath = XPathFactory.newInstance().newXPath();
        XPathExpression expr = xpath.compile("//text()[normalize-space()='']");
        NodeList nodeList = (NodeList)expr.evaluate(node, XPathConstants.NODESET);

        for (int i = 0; i < nodeList.getLength(); ++i) {
            Node nd = nodeList.item(i);
            nd.getParentNode().removeChild(nd);
        }

        // Create and setup transformer
        Transformer transformer =  TransformerFactory.newInstance().newTransformer();
        transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
        
        if (omitXmlDeclaration == true) {
           transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
        }

        if (prettyPrint == true) {
           transformer.setOutputProperty(OutputKeys.INDENT, "yes");
           transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
        }

        // Turn the node into a string
        StringWriter writer = new StringWriter();
        transformer.transform(new DOMSource(node), new StreamResult(writer));
        return writer.toString();
    } catch (TransformerException e) {
        throw new RuntimeException(e);
    } catch (XPathExpressionException e) {
        throw new RuntimeException(e);
    }
}

Answer 2

这是一种可能性：

void save(Node node, OutputStream stream) throws Exception {
    DOMImplementationLS ls = (DOMImplementationLS) node.getOwnerDocument().getImplementation();
    LSOutput out = ls.createLSOutput();
    out.setByteStream(stream);
    LSSerializer ser = ls.createLSSerializer();
    ser.write(node, out);
}

也许你应该确保node 真的是Element，但我把它留作excersize。