相关疑难解决方法(0)

我是XML的新手.我想根据请求名称阅读以下XML.请帮助我如何在Java中阅读以下XML -

<?xml version="1.0"?>
    <config>
        <Request name="ValidateEmailRequest">
            <requestqueue>emailrequest</requestqueue>
            <responsequeue>emailresponse</responsequeue>
        </Request>
        <Request name="CleanEmail">
            <requestqueue>Cleanrequest</requestqueue>
            <responsequeue>Cleanresponse</responsequeue>
        </Request>
    </config>

我想使用JDOM读取XML文件,然后使用XPath从JDOM文档中提取数据.它创建了Document对象,但是当我使用XPath查询Document的元素列表时,我什么也得不到.

我的XML文档在根元素中定义了一个默认命名空间.有趣的是,当我删除默认命名空间时,它成功运行XPath查询并返回我想要的元素.还有什么办法让我的XPath查询返回结果？

XML:

<?xml version="1.0" encoding="UTF-8"?>
<collection xmlns="http://www.foo.com">
<dvd id="A">
  <title>Lord of the Rings: The Fellowship of the Ring</title>
  <length>178</length>
  <actor>Ian Holm</actor>
  <actor>Elijah Wood</actor>
  <actor>Ian McKellen</actor>
</dvd>
<dvd id="B">
  <title>The Matrix</title>
  <length>136</length>
  <actor>Keanu Reeves</actor>
  <actor>Laurence Fishburne</actor>
</dvd>
</collection>

Java的:

public static void main(String args[]) throws Exception {
    SAXBuilder builder = new SAXBuilder();
    Document d = builder.build("xpath.xml");
    XPath xpath = XPath.newInstance("collection/dvd");
    xpath.addNamespace(d.getRootElement().getNamespace());
    System.out.println(xpath.selectNodes(d));
}

xml看起来像这样:

<statements>
   <statement account="123">
      ...stuff...
   </statement>
   <statement account="456">
      ...stuff...
   </statement>
</statements>

我正在使用stax一次处理一个" <statement>",然后我就开始工作了.我需要将整个语句节点作为字符串获取,这样我就可以创建"123.xml"和"456.xml",甚至可以将其加载到由account索引的数据库表中.

使用这种方法:http://www.devx.com/Java/Article/30298/1954

我想做这样的事情:

String statementXml = staxXmlReader.getNodeByName("statement");

//load statementXml into database

我知道这个页面有很多这个主题,但遗憾的是,我仍然无法得到我的解决方案..

这是我的xml代码:

<?xml version="1.0" encoding="UTF-8"?>
<ns1:Request xmlns:ns1="http://www.sea.com">
<ns1:PayrollRequest>
  <ns1:PayrollCost>
    <ns1:PayrollID>123</ns1:PayrollID>
    <ns1:BatchID>7770</ns1:BatchID>
    <ns1:CompanyId>001</ns1:CompanyId>
    <ns1:GrossPay>60000</ns1:GrossPay>
  </ns1:PayrollCost>
</ns1:PayrollRequest>
</ns1:Request>

这是我在java中的代码:

import org.w3c.dom.*;
import javax.xml.xpath.*;
import javax.xml.parsers.*;
import java.io.IOException;
import org.xml.sax.SAXException;

public class XPathTry {

public static void main(String[] args) 
throws ParserConfigurationException, SAXException, 
IOException, XPathExpressionException {

DocumentBuilderFactory domFactory = 
DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(true); 
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document doc = builder.parse("SamplePayroll2.xml");
XPath xpath = XPathFactory.newInstance().newXPath();


// display all
XPathExpression expr = xpath.compile("//PayrollCost/*/text()");


Object result = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodes = (NodeList) result;
for (int i = …

我有一个 XML 解析器StAX，我正在使用它来解析一个巨大的文件。但是，我想尽可能缩短时间。我正在读取将其放入数组中的值并将其发送到另一个函数进行评估。我正在调用该displayName标签，它应该在获取名称后立即转到下一个 xml，而不是读取整个 xml 文件。我正在寻找最快的方法。

爪哇：

import java.io.File;

import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.Iterator;
import javax.xml.namespace.QName;
import javax.xml.stream.XMLEventReader;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamException;
import javax.xml.stream.events.*;

public class Driver {

    private static boolean bname;

    public static void main(String[] args) throws FileNotFoundException, XMLStreamException {

        File file = new File("C:\\Users\\Robert\\Desktop\\root\\SDKCode\\src\\main\\java\\com\\example\\xmlClass\\data.xml");


        parser(file);
    }

    public static void parser(File file) throws FileNotFoundException, XMLStreamException {

        bname = false;


        XMLInputFactory factory = XMLInputFactory.newInstance();


        XMLEventReader eventReader = factory.createXMLEventReader(new FileReader(file));


        while (eventReader.hasNext()) {

            XMLEvent event = …