Java 迭代XPath以仅获取表名
我想迭代XPath以仅获取表名。这是我的Xpath值Java 迭代XPath以仅获取表名,java,xpath,Java,Xpath,我想迭代XPath以仅获取表名。这是我的Xpath值 PromoteData/Table[@Id='auditSet']/Table[@Id='auditSetMapping']/Table[@Id='appExpr'] 请指导我如何迭代此XPath的每个值。预期的输出类似于“auditSet”、“auditSetMapping”、“appExpr”import java.io.IOException; import java.io.IOException; import java.io.St
PromoteData/Table[@Id='auditSet']/Table[@Id='auditSetMapping']/Table[@Id='appExpr']
请指导我如何迭代此XPath的每个值。预期的输出类似于“auditSet”、“auditSetMapping”、“appExpr”import java.io.IOException;
import java.io.IOException;
import java.io.StringReader;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;
public class XPathExtract {
public static void main(String[] args) throws SAXException, IOException,
ParserConfigurationException, XPathExpressionException {
String xml = "...";
// Parse the XML as DOM
Document doc = DocumentBuilderFactory.newInstance().newDocumentBuilder()
.parse(new InputSource(new StringReader(xml)));
// Create a XPath instance
XPath xpath = XPathFactory.newInstance().newXPath();
// Evaluate the XPath. The result is a NodeList
NodeList nodes = (NodeList) xpath.evaluate(
"PromoteData/Table[@Id='auditSet']/Table[@Id='auditSetMapping']/Table[@Id='appExpr']",
doc, XPathConstants.NODESET);
// Iterate over the nodes
for (int i = 0; i < nodes.getLength(); i++) {
// node.item(i) is a Node. If you are sure, it is always an Element you can do a cast
Element el = (Element) nodes.item(i);
// Process the element
System.out.println(el.getAttribute("Id"));
}
}
}
导入java.io.StringReader;
导入javax.xml.parsers.DocumentBuilderFactory;
导入javax.xml.parsers.parserConfiguration异常;
导入javax.xml.xpath.xpath;
导入javax.xml.xpath.XPathConstants;
导入javax.xml.xpath.XPathExpressionException;
导入javax.xml.xpath.XPathFactory;
导入org.w3c.dom.Document;
导入org.w3c.dom.Element;
导入org.w3c.dom.NodeList;
导入org.xml.sax.InputSource;
导入org.xml.sax.SAXException;
公共类XPathExtract{
公共静态void main(字符串[]args)引发SAXException、IOException、,
ParserConfiguration异常,XPathExpressionException{
字符串xml=“…”;
//将XML解析为DOM
Document doc=DocumentBuilderFactory.newInstance().newDocumentBuilder()
.parse(新的InputSource(新的StringReader(xml));
//创建一个XPath实例
XPath=XPathFactory.newInstance().newXPath();
//计算XPath。结果是NodeList
NodeList节点=(NodeList)xpath.evaluate(
“PromoteData/Table[@Id='auditSet']/Table[@Id='auditSetMapping']/Table[@Id='appExpr']”,
doc,XPathConstants.NODESET);
//在节点上迭代
对于(int i=0;i
复制自为什么坚持编辑示例xpath值?没有它,这个问题就没有多大意义。对于任何其他人来说,这个问题都可以在编辑历史记录中找到。删除这个帖子没有任何依据,对帮助你的人甚至想这么做都是不公平的。编辑代码也是不合适的。我把它还给你了。不要再这样做了。谢谢vanje,但这段代码只打印了1个值,即appExprr,我希望输出为auditset、auditsetmapping、appExprMaybe。您应该调整XPath表达式。但是您没有提供XML示例,因此很难检查XPath表达式是否符合您的需要。