将xpath的结果转换为xml格式答案

【问题标题】：transform result of xpath to xml format将xpath的结果转换为xml格式
【发布时间】：2016-03-14 08:39:57
【问题描述】：

我的 xml 文件如下：

<?xml version="1.0"?>
<all>
<test1>
hajarrrr rrr rr
</test1>
<catalog>
   <book id="bk101">
      <author>
<infos>Empire Burlesque</infos> 
</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>
<infos>Emhhsjshhh</infos>  
</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, and her own childhood to become queen 
      of the world.</description>
   </book>
</catalog>
</all>

我只想提取 <catalog> 和 </catalog> 之间的块，所以我写了这个 java 代码：

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

        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = factory.newDocumentBuilder();
        Document doc = builder.parse("C:\\Users\\HC\\Desktop\\dataset\\book.xml");
        XPathFactory xPathfactory = XPathFactory.newInstance();
        XPath xpath = xPathfactory.newXPath();
        //TODO: Is this correct query?
        XPathExpression xpathExp = xpath.compile("//text()[normalize-space(.) = '']");
        NodeList emptyTextNodes = (NodeList) xpathExp.evaluate(doc, XPathConstants.NODESET);
        for (int i = 0; i < emptyTextNodes.getLength(); i++) {
            Node emptyTextNode = emptyTextNodes.item(i);
            emptyTextNode.getParentNode().removeChild(emptyTextNode);
        }

        XPathExpression expr = xpath.compile("/all/catalog/descendant::node()");
        NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
        for (int i = 0; i < nl.getLength(); i++) {
            Node n = nl.item(i);

          System.out.println(n.getNodeName() + "   " + n.getNodeValue());

        }
    }

我希望结果是 xml 格式，而不是这样：

book   null
author   null
infos   null
#text   Empire Burlesque
title   null
#text   XML Developer's Guide
genre   null
#text   Computer
price   null
#text   44.95
publish_date   null
#text   2000-10-01
description   null
#text   An in-depth look at creating applications 
      with XML.
book   null
author   null
infos   null
#text   Emhhsjshhh
title   null
#text   Midnight Rain
genre   null
#text   Fantasy
price   null
#text   5.95
publish_date   null
#text   2000-12-16
description   null
#text   A former architect battles corporate zombies, 
      an evil sorceress, and her own childhood to become queen 
      of the world.

事实上，我想要这样的结果：

 <book id="bk101">
      <author>
<infos>Empire Burlesque</infos> 
</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>
<infos>Emhhsjshhh</infos>  
</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, and her own childhood to become queen 
      of the world.</description>
   </book>

请任何人帮忙:)，并在此先感谢。

【问题讨论】：

XML Node to String in Java
你已经用 JDOM 标记了你的问题，但它不是 JDOM，它只是 DOM。 JDOM 将使整个事物看起来大不相同，而且可能更加自然。您应该使用 DOM，还是应该使用 JDOM？ DOM tutorial 和 JDOM tutorial
这是 dom :D 抱歉！

标签： java xml dom xpath

【解决方案1】：

¿典型的文件解析器怎么样？

File file = new File ("xml_path");
BufferedReader br = new BufferedReader (new FileReader (file));
String info;

while (info = br.readLine() != null)
{
    if (info.contains("<catalog>"))
    {
        System.out.println(info);
        while (info = br.readLine() != null)
        {
            if (info.contains("</catalog>"))
            {
                 System.out.println(info);
                 break;
            }
            else System.out.println(info);
        }
    }
}

这只是一个简单的文本解析器。不知道有没有更好的办法，希望对你有帮助:)

【讨论】：