Java xpath以字符串形式返回整个元素

时间:2012-12-19 01:20:18

标签: java xml xpath

我需要使用java xpath将id和xml元素作为字符串返回。

...给出

<svg>
    <g id="Background">
    </g>
    <g id="Outline">
        <polygon fill="none" stroke="#000000" stroke-linecap="round" stroke-linejoin="round"     stroke-miterlimit="10" points=" 119.813,57.875 119.188,57.87" />
    </g>
    <g id="Base_Colour" transform="matrix(0.25 0 0 0.25 0 0)">
        <path fill="#ADB1AF" d="M112.25,208l-8,20.25l-0.5-1.75l0.75-0.5v-1.5l0.75-0.5v-1.5L106,222v-1.5l0.75-0.5v-1.5l0.75-0.5v-1.5"/>
        <path fill="#625595" d="M112.25,208l5.25-14.5l30-30.25l2.25-1.5l41.5-20.5l49.75-9.5h4.25l49,3l48.75"/>
    </g>
</svg>

返回的值必须是......

<g id="Outline">
    <polygon fill="none" stroke="#000000" stroke-linecap="round" stroke-linejoin="round"     stroke-miterlimit="10" points=" 119.813,57.875 119.188,57.87" />
</g> 

我已经广泛搜索过,我尝试过的任何东西都无法返回整个元素。需要Xpath,因为我想通过id查询任何级别的g标签。

5 个答案:

答案 0 :(得分:8)

我找到的解决方案是使用xpath获取org.w3c.dom.Node(DOM也可以)。然后我从节点创建了一个javax.xml.transform.dom.DOMSource,并将其转换为带有javax.xml.transform.TransformerFactory的字符串。

Node node = // the node you want to serialize
xmlOutput = new StreamResult(new StringWriter());
transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.transform(new DOMSource(node), xmlOutput);
String nodeAsAString = xmlOutput.getWriter().toString();

这很容易被考虑到一个类中以供重用。不幸的是,Java中没有.OuterXml属性,就像在.NET中一样。你们所有的.NETer现在都可以傻笑。

答案 1 :(得分:1)

没有xpath会返回包含XML语法的字符串。

答案 2 :(得分:0)

我用这段代码解决了我的问题:

public static String getOuterXml(Node node)
    throws TransformerConfigurationException, TransformerException {
    Transformer transformer = TransformerFactory.newInstance().newTransformer();
    transformer.setOutputProperty("omit-xml-declaration", "yes");

    StringWriter writer = new StringWriter();
    transformer.transform(new DOMSource(node), new StreamResult(writer));
    return writer.toString();         
}

致记:chick.Net

答案 3 :(得分:0)

有时候,您必须在Java中没有xml文档的情况下执行此操作;我发现下面的代码非常用全

import java.io.IOException;
import java.io.StringReader;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
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 org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;

String responseMsg = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><YourMessage><MyTag>MyTagValue</MyTag></YourMessage>";
String expressionToExract = "/YourMessage/MyTag";
String xmlNodeWithData = xpathTester.getXmlNode(responseMsg, expressionToExract);
//above xmlNodeWithData will have this value '<MyTag>MyTagValue</MyTag>'

private String getXmlNode(String resultMsg, String expression)
        throws ParserConfigurationException, SAXException, IOException, XPathExpressionException {
    String xmlNodeWithData="";
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder = factory.newDocumentBuilder();
    InputSource is = null;
    StringReader sr = null;
    sr = new StringReader(resultMsg);
    is = new InputSource(sr);

    Document doc = builder.parse(is);
    XPathFactory xPathfactory = XPathFactory.newInstance();
    XPath xpath = xPathfactory.newXPath();
    XPathExpression expr = xpath.compile(expression);

    Node node = (Node)expr.evaluate(doc, XPathConstants.NODE);      
    xmlNodeWithData += "<" + node.getNodeName() + ">";
    NodeList nodeList = node.getChildNodes();

    for (int nodeIndex=0; nodeIndex < nodeList.getLength(); nodeIndex++) {
        Node nodeChild = nodeList.item(nodeIndex);          
        if (nodeChild.getNodeName().contains("#text")) {
            xmlNodeWithData += nodeChild.getTextContent();
            continue;
        }
        xmlNodeWithData += "<" + nodeChild.getNodeName() + ">";         
        xmlNodeWithData += nodeChild.getTextContent();
        xmlNodeWithData += "</" + nodeChild.getNodeName() + ">";
    }
    xmlNodeWithData += "</" + node.getNodeName() + ">";
    if (sr != null) {
        sr.close();
    }
    return xmlNodeWithData;
}

答案 4 :(得分:-1)

我不了解Java,但在.NET世界中我们会使用

doc.DocumentElement.SelectSingleNode("/*/g[@id='Outline']").OuterXml