Question

我的目标是使用XPath执行XQuery。

我的XML文件是：

<?xml version="1.0" encoding="UTF-8"?>
<postes>
    <poste>
        <gouvernourat>Kairouan</gouvernourat>
        <ville>Kairouan sud</ville>
        <cp>3100</cp>
    </poste>
    <poste>
        <gouvernourat>Tunis</gouvernourat>
        <ville>Ghazela</ville>
        <cp>1002</cp>
    </poste>
</postes>

我的Java代码是：

package xmlparse;


import java.io.IOException;
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.NodeList;
import org.xml.sax.SAXException;

public class QueryXML {
    public void query() throws ParserConfigurationException, SAXException,
            IOException, XPathExpressionException {
        // Standard of reading a XML file
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true);
        DocumentBuilder builder;
        Document doc = null;
        XPathExpression expr = null;
        builder = factory.newDocumentBuilder();
        doc = builder.parse("a.xml");  //C:\\Users\\aymen\\Desktop\\

        // Create a XPathFactory
        XPathFactory xFactory = XPathFactory.newInstance();

        // Create a XPath object
        XPath xpath = xFactory.newXPath();

        // Compile the XPath expression
        expr = xpath.compile("/postes/poste[gouvernourat='Tunis']/ville/text()");
        // Run the query and get a nodeset
        Object result = expr.evaluate(doc, XPathConstants.NODESET);

        // Cast the result to a DOM NodeList
        NodeList nodes = (NodeList) result;
        for (int i=0; i<nodes.getLength();i++){
            System.out.println(nodes.item(i).getNodeValue());
        }
    }

    public static void main(String[] args) throws XPathExpressionException, ParserConfigurationException, SAXException, IOException {
        QueryXML process = new QueryXML();
        process.query();
    }
}

当我启动此Java代码时，结果会在控制台上正确显示(System.out.println)。

但是，如果我将此代码复制到我的Android应用程序并将System.out.println(nodes.item(i).getNodeValue());更改为Text2.setText(nodes.item(i).getNodeValue());（我有一个名为Text2的TextView）

当我执行代码并单击按钮时，TextView保持为空（Force Close没有错误）

提前谢谢

Answer 1

在Android中使用XPath时，属性名称需要以“@”开头。
所以改变

[gouvernourat='Tunis']

要

[@gouvernourat='Tunis']

有关详细信息，请参阅http://developer.android.com/reference/javax/xml/xpath/package-summary.html。

Android中的XPath错误

1 个答案: