我需要使用NodeList创建一个
XML Document对象.有人可以帮我做这件事.我已经向您展示了下面的代码和xml
import
javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.*; import
org.w3c.dom.*;
public class ReadFile {
public static void main(String[] args) {
String exp = "/configs/markets";
String path = "testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList)
xPathExpression.evaluate(xmlDocument,XPathConstants.NODESET);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
xml文件如下所示
<configs>
<markets>
<market>
<name>Real</name>
</market>
<market>
<name>play</name>
</market>
</markets>
</configs>
提前致谢..
解决方法
你应该这样做：
>您创建一个新的org.w3c.dom.Document newXmlDoc,其中存储NodeList中的节点,@H_502_15@>您创建一个新的根元素,并将其附加到newXmlDoc@H_502_15@>然后,对于NodeList中的每个节点n,在newXmlDoc中导入n,然后将n附加为root的子节点
这是代码：
public static void main(String[] args) {
String exp = "/configs/markets/market";
String path = "src/a/testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList) xPathExpression.
evaluate(xmlDocument,XPathConstants.NODESET);
Document newXmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().newDocument();
Element root = newXmlDocument.createElement("root");
newXmlDocument.appendChild(root);
for (int i = 0; i < nodes.getLength(); i++) {
Node node = nodes.item(i);
Node copyNode = newXmlDocument.importNode(node,true);
root.appendChild(copyNode);
}
printTree(newXmlDocument);
} catch (Exception ex) {
ex.printStackTrace();
}
}
public static void printXmlDocument(Document document) {
DOMImplementationLS domImplementationLS =
(DOMImplementationLS) document.getImplementation();
LSSerializer lsSerializer =
domImplementationLS.createLSSerializer();
String string = lsSerializer.writeToString(document);
System.out.println(string);
}
输出是：
<?xml version="1.0" encoding="UTF-16"?>
<root><market>
<name>Real</name>
</market><market>
<name>play</name>
</market></root>
一些说明：
>我已经将exp改为/ configs / markets / market,因为我怀疑你想要复制市场元素,而不是单一的市场元素@H_502_15@>对于printXmlDocument,我在这个answer中使用了有趣的代码
我希望这有帮助.
如果您不想创建新的根元素,那么您可以使用原始的XPath表达式,它返回一个由单个节点组成的NodeList(请记住,您的XML必须只有一个根元素),您可以直接添加到您的新XML文档.
public static void main(String[] args) {
//String exp = "/configs/markets/market/";
String exp = "/configs/markets";
String path = "src/a/testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList) xPathExpression.
evaluate(xmlDocument,XPathConstants.NODESET);
Document newXmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().newDocument();
//Element root = newXmlDocument.createElement("root");
//newXmlDocument.appendChild(root);
for (int i = 0; i < nodes.getLength(); i++) {
Node node = nodes.item(i);
Node copyNode = newXmlDocument.importNode(node,true);
newXmlDocument.appendChild(copyNode);
//root.appendChild(copyNode);
}
printXmlDocument(newXmlDocument);
} catch (Exception ex) {
ex.printStackTrace();
}
}
这将为您提供以下输出：
<?xml version="1.0" encoding="UTF-16"?>
<markets>
<market>
<name>Real</name>
</market>
<market>
<name>play</name>
</market>
</markets>
