如何提取XML的某些部分，而不是具有某些不相似性的部分
如何提取XML的某些部分，而不是具有某些不相似性的部分
提问于 2019-06-14 01:41:47
所以我有一个包含块的XML文件，我需要提取其中一些价格高于10，发布日期高于2005年的块。
我刚用getElementById把所有的模块都拉出来了。我在想，如果不是这样的话..。
public class Helper
{
public static void parseXML()
{
try {
File inputFile = new File("XML.xml");
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(inputFile);
doc.getDocumentElement().normalize();
System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
NodeList nList = doc.getElementsByTagName("book");
System.out.println("----------------------------");
for (int temp = 0; temp < nList.getLength(); temp++) {
Node nNode = nList.item(temp);
System.out.println("\nCurrent Element :" + nNode.getNodeName());
if (nNode.getNodeType() == Node.ELEMENT_NODE) {
Element eElement = (Element) nNode;
System.out.println("First Name : "
+ eElement
.getElementsByTagName("author")
.item(0)
.getTextContent());
System.out.println("Title : "
+ eElement
.getElementsByTagName("title")
.item(0)
.getTextContent());
System.out.println("Genre : "
+ eElement
.getElementsByTagName("genre")
.item(0)
.getTextContent());
System.out.println("Price : "
+ eElement
.getElementsByTagName("price")
.item(0)
.getTextContent());
System.out.println("Publish date : "
+ eElement
.getElementsByTagName("publish_date")
.item(0)
.getTextContent());
System.out.println("Description : "
+ eElement
.getElementsByTagName("description")
.item(0)
.getTextContent());
}
}
}
catch (SAXException ex) { }
catch (IOException ex) { }
catch (ParserConfigurationException ex) { }
}
}<?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>2002-12-16</publish_date>
<description>A former architect battles corporate zombies,
an evil sorceress, and her own childhood to become queen
of the world.</description>
</book>
<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, the young survivors lay the
foundation for a new society.</description>
</book>
<book id="bk104">
<author>Corets, Eva</author>
<title>Oberon's Legacy</title>
<genre>Fantasy</genre>
<price>5.95</price>
<publish_date>2001-03-10</publish_date>
<description>In post-apocalypse England, the mysterious
agent known only as Oberon helps to create a new life
for the inhabitants of London. Sequel to Maeve
Ascendant.</description>
</book>
<book id="bk105">
<author>Corets, Eva</author>
<title>The Sundered Grail</title>
<genre>Fantasy</genre>
<price>5.95</price>
<publish_date>2001-09-10</publish_date>
<description>The two daughters of Maeve, half-sisters,
battle one another for control of England. Sequel to
Oberon's Legacy.</description>
</book>
<book id="bk106">
<author>Randall, Cynthia</author>
<title>Lover Birds</title>
<genre>Romance</genre>
<price>4.95</price>
<publish_date>2003-09-02</publish_date>
<description>When Carla meets Paul at an ornithology
conference, tempers fly as feathers get ruffled.</description>
</book>
<book id="bk107">
<author>Thurman, Paula</author>
<title>Splish Splash</title>
<genre>Romance</genre>
<price>4.95</price>
<publish_date>2004-11-02</publish_date>
<description>A deep sea diver finds true love twenty
thousand leagues beneath the sea.</description>
</book>
<book id="bk108">
<author>Knorr, Stefan</author>
<title>Creepy Crawlies</title>
<genre>Horror</genre>
<price>4.95</price>
<publish_date>2005-12-06</publish_date>
<description>An anthology of horror stories about roaches,
centipedes, scorpions and other insects.</description>
</book>
<book id="bk109">
<author>Kress, Peter</author>
<title>Paradox Lost</title>
<genre>Science Fiction</genre>
<price>6.95</price>
<publish_date>2006-11-02</publish_date>
<description>After an inadvertant trip through a Heisenberg
Uncertainty Device, James Salway discovers the problems
of being quantum.</description>
</book>
<book id="bk110">
<author>O'Brien, Tim</author>
<title>Microsoft .NET: The Programming Bible</title>
<genre>Computer</genre>
<price>36.95</price>
<publish_date>2006-12-09</publish_date>
<description>Microsoft's .NET initiative is explored in
detail in this deep programmer's reference.</description>
</book>
<book id="bk111">
<author>O'Brien, Tim</author>
<title>MSXML3: A Comprehensive Guide</title>
<genre>Computer</genre>
<price>36.95</price>
<publish_date>2007-12-01</publish_date>
<description>The Microsoft MSXML3 parser is covered in
detail, with attention to XML DOM interfaces, XSLT processing,
SAX and more.</description>
</book>
<book id="bk112">
<author>Galos, Mike</author>
<title>Visual Studio 7: A Comprehensive Guide</title>
<genre>Computer</genre>
<price>49.95</price>
<publish_date>2008-04-16</publish_date>
<description>Microsoft Visual Studio 7 is explored in depth,
looking at how Visual Basic, Visual C++, C#, and ASP+ are
integrated into a comprehensive development
environment.</description>
</book>
</catalog>我希望输出只是价格更高为10的块。我试图将字符串(价格为getElementById)解析为整数并声明它，然后将其放入，如果像这样( if (a>10))，代码中没有错误，但输出是错误的。它说Integer是不好的。
回答 1
Stack Overflow用户
发布于 2019-06-14 03:13:13
嗯，如果这是家庭作业，它可能会对解决方案有一些限制。但是，查询xml文档的正确工具是XPath。它是一种灵活的查询语言，解析器是JDK的一部分。
下面是一个解决赋值问题的示例
try {
Document document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(
Files.newInputStream(Paths.get("C:/temp/xx.xml")));
XPath xPath = XPathFactory.newInstance().newXPath();
// find elements in path /catalog/book where price element > 10.00 and first 4 letters of element publish_date are greater than '2005'
String expression = "/catalog/book[price > 10.00 and substring(publish_date,1,4) > '2005']";
NodeList list = (NodeList) xPath.compile(expression).evaluate(document, XPathConstants.NODESET);
for (int i = 0 ; i < list.getLength() ; i++) {
Node node = list.item(i);
String id = node.getAttributes().getNamedItem("id").getTextContent();
System.out.println(id);
}
} catch (Exception e) {
e.printStackTrace();
}输出
bk110
bk111
bk112页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56585934
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