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XPATH 解析 xml 的更好方法

转载 作者:行者123 更新时间:2023-12-02 00:26:23 25 4
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解析此类 xml 的更好方法是什么:

<FindLicensesResponse xmlns="http://abc.com">
<FindLicensesResult>
<Licensies>
<ActivityLicense>
<id>1</id>
<DateIssue>2011-12-29T00:00:00</DateIssue>
<ActivityType xmlns:s01="http://www.w3.org/2001/XMLSchema-instance" s01:type="ActivityType">
<code>somecode1</code>
</ActivityType>
<ActivityTerritory xmlns:s02="http://www.w3.org/2001/XMLSchema-instance" s02:type="Territory">
<code>somecode2</code>
</ActivityTerritory>
<ActivityLicenseAttachments />
</ActivityLicense>
<ActivityLicense>
<id>2</id>
<DateIssue>2011-12-21T00:00:00</DateIssue>
<ActivityType xmlns:s01="http://www.w3.org/2001/XMLSchema-instance" s01:type="ActivityType">
<code>somecode3</code>
</ActivityType>
<ActivityTerritory xmlns:s02="http://www.w3.org/2001/XMLSchema-instance" s02:type="Territory">
<code>somecode4</code>
</ActivityTerritory>
<ActivityLicenseAttachments />
</ActivityLicense>
</Licensies>
</FindLicensesResult>

我需要从每个 ActivityLicense: id、DateIssue 和内部 ActivityType: code 和内部 ActivityTerritory: code 获取值。

现在我这样做:

CachedXPathAPI xpathAPI = new CachedXPathAPI();
Element nsctx = result.getSOAPPart().createElementNS(null, "nsctx");
nsctx.setAttributeNS("http://www.w3.org/2000/xmlns/","xmlns:el","http://abc.com");
NodeList activityLicenses = xpathAPI.selectNodeList(result.getSOAPPart(),"//el:ActivityLicense", nsctx);

for (int i = 0; i < activityLicenses.getLength(); i++) {
Node id = xpathAPI.selectSingleNode(activityLicenses.item(i), "//el:id", nsctx);
Node dateIssue = xpathAPI.selectSingleNode(activityLicenses.item(i), "//el:DateIssue",nsctx);

System.out.println("id: " + id.getTextContent());
System.out.println("dateIssue: " + dateIssue.getTextContent());
}

但是我无法从 ActivityType/code 和 ActivityTerritory/code 获取值

最佳答案

看看这个解决方案

import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.InputStream;

import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;

public class StringTest {

public static void main(String[] args) throws Exception {
String xml = "";
java.util.Scanner sc = new java.util.Scanner(new File("xml.xml"));
while(sc.hasNextLine()){
xml+=sc.nextLine();
}
javax.xml.parsers.DocumentBuilderFactory dbFactory = javax.xml.parsers.DocumentBuilderFactory.newInstance();
javax.xml.parsers.DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
InputStream is = new ByteArrayInputStream(xml.getBytes());
org.w3c.dom.Document doc = dBuilder.parse(is);
doc.getDocumentElement().normalize();
XPath xpath = XPathFactory.newInstance().newXPath();

org.w3c.dom.NodeList nodeList = doc.getElementsByTagName("ActivityLicense");
for(int i=0;i<nodeList.getLength();i++){
org.w3c.dom.Node node = nodeList.item(i);
System.out.println(xpath.evaluate("ActivityTerritory/code/text()", node, XPathConstants.STRING));
}
}
}

关于XPATH 解析 xml 的更好方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14972569/

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