如何获取特定的XML元素参数值？

Question

我开始解析xml文档并有疑问:如何在Java上获取特定的XML元素参数值？

XML文档:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<data>
<keyword name="text123">
    <profile num="1">
        <url>http://www.a.com</url>
        <field-1 param="">text</field-1>
        <filed-2 param="">text</field-2>
    </profile>
    <profile num="2">
        <url>http://www.b.com</url> 
        <field-1 param="">text</field-1>
        <filed-2 param="">text</field-2>
    </profile>
</keyword>
<keyword name="textabc123">
    <profile num="1">
        <url>http://www.1a.com</url>
        <field-1 param="">text</field-1>
        <filed-2 param="">text</field-2>
    </profile>
    <profile num="2">
        <url>http://www.1b.com</url> 
        <field-1 param="">text</field-1>
        <filed-2 param="">text</field-2>
    </profile>
</keyword>
</data>

我在Java上写的代码:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
File xml_file=new File("file.xml");
if (xml_file.isFile() && xml_file.canRead()) {
 Document doc = builder.parse(xml_file);
 Element root = doc.getDocumentElement();
 NodeList nodel = root.getChildNodes();
 for (int a = 0; a < nodel.getLength(); a++) {
  String data = /* code i don't know to write*/
  System.out.println(data);
 }
} else {}

我想出去控制台元素"关键字"参数"名称"值:

text123

和

text123abc

请帮忙,谢谢.

Answer 1

您可以使用XPath

InputStream is = getClass().getResourceAsStream("somefile.xml");
DocumentBuilderFactory xmlFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = xmlFactory.newDocumentBuilder();
Document xmlDoc = docBuilder.parse(is);
XPathFactory xpathFact = XPathFactory.newInstance();
XPath xpath = xpathFact.newXPath();

String text123 = (String) xpath.evaluate("/data/keyword[1]/@name", xmlDoc, XPathConstants.STRING);
String textabc123 = (String) xpath.evaluate("/data/keyword[2]/@name", xmlDoc, XPathConstants.STRING);

Answer 2

我将指导您如何使用JAXB来完成它.

首先,您的XML格式不正确.你有filed而不是field在几个地方.

适当的XML:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<data>
<keyword name="text123">
    <profile num="1">
        <url>http://www.a.com</url>
        <field-1 param="">text</field-1>
        <field-2 param="">text</field-2>
    </profile>
    <profile num="2">
        <url>http://www.b.com</url> 
        <field-1 param="">text</field-1>
        <field-2 param="">text</field-2>
    </profile>
</keyword>
<keyword name="textabc123">
    <profile num="1">
        <url>http://www.1a.com</url>
        <field-1 param="">text</field-1>
        <field-2 param="">text</field-2>
    </profile>
    <profile num="2">
        <url>http://www.1b.com</url> 
        <field-1 param="">text</field-1>
        <field-2 param="">text</field-2>
    </profile>
</keyword>
</data>

接下来,访问该网站并下载Trang.

假设您的XML文件已命名sample.xml,请通过Trang运行它java -jar trang.jar sample.xml sample.xsd以获取xml文件的xsd架构.

现在,运行xjc sample.xsd(xjc是一个用于为XML模式生成Java类的工具,它与Java 6 SDK捆绑在一起).

您将获得一个Java类列表:

generated\Data.java
generated\Field1.java
generated\Field2.java
generated\Keyword.java
generated\ObjectFactory.java
generated\Profile.java

将它们放在Java项目文件中,放在sample.xml程序可以找到它的位置.现在,这是获取关键字名称的方式:

JAXBContext context = JAXBContext.newInstance(Data.class);
Data data = (Data)context.createUnmarshaller().unmarshal(new File("sample.xml"));

List<Keyword> keywords = data.getKeyword();

for (Keyword keyword : keywords) {
    System.out.println(keyword.getName());
}

这个方法在开始时可能看起来有点混乱,但是如果你的XML结构没有改变,我发现处理类型化Java对象比使用DOM本身更好.