用于简单 XML 节点字符串的 Android XML 解析器或库

Question

这是一个 XML 文件的例子:

  <?xml version="1.0"?>
  <catalog>
    <book id="bk101">
     <author>Gambardella, Matthew</author>
     <title>XML Developer's Guide</title>
     <genre>Computer</genre>
     <price>44.95</price>
     <publish_date>2000-10-01</publish_date>
     <description>An in-depth look at creating applications 
      with XML.</description>
    </book>

    <book id="bk102">
     <author>Ralls, Kim</author>
     <title>Midnight Rain</title>
     <genre>Fantasy</genre>
     <price>5.95</price>
     <publish_date>2000-12-16</publish_date>
     <description>A former architect battles corporate zombies, 
      an evil sorceress.</description>
    </book>

    <book id="bk103">
     <author>Corets, Eva</author>
     <title>Maeve Ascendant</title>
     <genre>Fantasy</genre>
     <price>5.95</price>
     <publish_date>2000-11-17</publish_date>
     <description>After the collapse of a nanotechnology 
      society in England.</description>
     </book>
  </catalog>

Si 我想按许多标准在此文件中搜索书籍，例如按作者、按流派、价格等...

我将使用 XPath 查询来执行此操作，那么有一些简单的方法可以使用吗？？？例如，我想检查作者是否存在，为此我必须有一个方法，我将在其中传递一个 XPath 查询来为我提供结果...

提前致谢

最好的问候，

阿里

Answer 1

您可以将 DOM API 与 javax.xml.xpath.XPathFactory 结合使用和 javax.xml.xpath.XPath如 http://developer.android.com/reference/javax/xml/xpath/package-summary.html 所述.

例如:

DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();;
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document document = builder.parse("input.xml");

// NodeList books = document.getElementsByTagName("book");

XPath xpath = XPathFactory.newInstance().newXPath();
String expression = "/catalog/book[1]/author"; // first book
Node author = (Node) xpath.evaluate(expression, document, XPathConstants.NODE);

if (author == null)
    System.out.println("Element author not exists");
else
    System.out.println(author.getTextContent());