Java XML Parse / Query

Question

当我使用NodeList时，我有这样的XML结构nList = doc.getElementsByTagName（＆＃34; stock＆＃34;）;它归还给我3个股票，2个主要股票标签和一个在亚组件下。我想只获得两个位于较高级别的股票，并忽略所有在子标签下的股票。

在Java中是否有可能在C＃中创建LINQ查询之类的东西，比如只返回名称等于＆＃34;索尼＆＃34;的元素。

谢谢！

<city>
   <stock>
     <name>Sony</name>
   </stock>
   <stock>
     <name>Panasonic</name>
     <substocks>
          <stock>
             <name>Panasonic Shop 2</name>
          </stock>
     </substocks>
   </stock>
</city>

Answer 1

我建议您将XPath与javax.xml.xpath包一起使用：

final InputStream is = new FileInputStream('your.xml');

final DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
final DocumentBuilder builder = factory.newDocumentBuilder();
final Document doc = builder.parse(is);
final XPathFactory xPathfactory = XPathFactory.newInstance();
final XPath xpath = xPathfactory.newXPath();
final XPathExpression expr = xpath.compile("/city/stock/name[text()='Sony']");

然后：

final NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);

Answer 2

看看XPath及其java实现JXPath。其他可能的方法是使用LambdaJ使用JAXB和操作对象列表解析XML。

Answer 3

还有dom4j库，它具有强大的XPath导航功能：

import org.dom4j.Document;
import org.dom4j.io.SAXReader;

SAXReader reader = new SAXReader();
Document document = reader.read("test.xml");
List list = document.selectNodes("/city/stock/name[text()='Sony']");
for (Iterator iter = list.iterator(); iter.hasNext(); ) {
    // TODO: place you logic here
}

更多示例是here

Answer 4

尝试jcabi-xml（请参阅此blog post）并使用一行代码

Collection<XML> found = new XMLDocument("your document here").nodes(
  "/city/stock/name[text()='Sony']"
);