java - 除了为此使用 Xpath 之外还有其他方法吗？

Question

大家好，我正在写这个程序:

import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NodeList;

public class DOMbooks {
   public static void main(String[] args) throws Exception {
      DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
      DocumentBuilder docBuilder = factory.newDocumentBuilder();
      File file = new File("books-fixed.xml");
      Document doc = docBuilder.parse(file);
      NodeList list = doc.getElementsByTagName("*");
      int bookCounter = 1;
      for (int i = 1; i < list.getLength(); i++) {
         Element element = (Element)list.item(i);
         String nodeName = element.getNodeName();
         if (nodeName.equals("book")) {
            bookCounter++;
            System.out.println("BOOK " + bookCounter);
            String isbn = element.getAttribute("sequence");
            System.out.println("\tsequence:\t" + isbn);
         } 
         else if (nodeName.equals("author")) {
            System.out.println("\tAuthor:\t" + element.getChildNodes().item(0).getNodeValue());
         }
         else if (nodeName.equals("title")) {
            System.out.println("\tTitle:\t" + element.getChildNodes().item(0).getNodeValue());
         } 
         else if (nodeName.equals("publishYear")) {
            System.out.println("\tpublishYear:\t" + element.getChildNodes().item(0).getNodeValue());
         } 
         else if (nodeName.equals("genre")) {
            System.out.println("\tgenre:\t" + element.getChildNodes().item(0).getNodeValue());
         } 
      }
   }
}

我想打印有关“科幻小说”书籍的所有数据。我知道我应该使用 Xpath，但它卡住了，错误太多……有什么建议吗？假设我有这张表，我只想选择包含所有信息的科幻小说

 <book sequence="5">
  <title>Aftershock</title> 
  <auther>Robert B. Reich</auther> 
  <publishYear>2010</publishYear> 
  <genre>Economics</genre> 
  </book>
- <book sequence="6">
  <title>The Time Machine</title> 
  <auther>H.G. Wells</auther> 
  <publishYear>1895</publishYear> 
  <genre>Science Fiction</genre> 

假设我有这张表，我只想打印科幻小说及其所有信息...

Answer 1

i want to print all the data about the "Science Fiction" books.. i know i should use Xpath but it's stuck,

我假设您的意思是您想要所有 genre == "Science Fiction" 的书，对吗？在这种情况下，XPath 确实比您在 Java 中尝试的任何方法都简单得多(您不显示根注释，所以我将从“//”开始，它可以在任何深度进行选择):

//book[genre = 'Science Fiction']

简化事情的 XSLT 方法

现在，再看一下您的代码，您似乎想要打印每个元素，包括元素的名称。这在 XSLT 中更简单:

<!-- every XSLT 1.0 must start like this -->
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

    <!-- you want text -->
    <xsl:output method="text" />

    <!-- match any science fiction book (your primary goal) -->
    <xsl:template match="book[genre = 'Science Fiction']">

        <xsl:text>BOOK </xsl:text>
        <xsl:value-of select="position()" />

        <!-- send the children and attribute to be processed by templates -->
        <xsl:apply-templates select="@sequence | *" />
    </xsl:template>

    <!-- "catch" any elements or attributes under <book> -->
    <xsl:template match="book/* | book/@*">

        <!-- a newline and a tab per line-->
        <xsl:text>&#xA;&#9;</xsl:text>

        <!-- and the name of the element or attribute -->
        <xsl:value-of select="local-name()" />

        <!-- another tab, plus contents of the element or attribute -->
        <xsl:text>&#9;</xsl:text>
        <xsl:value-of select="." />
    </xsl:template>

    <!-- make sure that other values are ignored, but process children -->
    <xsl:template match="node()">
        <xsl:apply-templates />
    </xsl:template>

</xsl:stylesheet>

您可以使用这段代码，它比您的原始代码更短(如果您忽略注释和空格)并且(可以说，一旦您掌握了它)更易读。要使用它:

• 将其存储为 books.xsl

• 然后，只需使用此 ( copied and changed from here):

  import javax.xml.transform.*;
  import javax.xml.transform.stream.StreamResult;
  import javax.xml.transform.stream.StreamSource;
  import java.io.File;
  import java.io.IOException;
  import java.net.URISyntaxException;
  
  public class TestMain {
      public static void main(String[] args) throws IOException, URISyntaxException, TransformerException {
          TransformerFactory factory = TransformerFactory.newInstance();
          Source xslt = new StreamSource(new File("books.xsl"));
          Transformer transformer = factory.newTransformer(xslt);
  
          Source text = new StreamSource(new File("books-fixed.xml"));
          transformer.transform(text, new StreamResult(new File("output.txt")));
      }
  }
  

XPath 2.0

如果可以使用Saxon在 Java 中，以上内容成为 XPath 2.0 的单行代码，您甚至不需要 XSLT:

for $book in //book[genre = 'Science Fiction']
return (
    'BOOK', 
    count(//book[genre = 'Science Fiction'][. << $book]) + 1,
    for $tag in $book/(@sequence | *)
    return $tag/local-name(), ':', string($tag)
)