java - xpath 获取子节点相对于 XML 中父节点的位置

Question

我需要循环遍历父节点(callEvents)中的所有子节点，并每次保存子节点的位置。(第一次调用事件，第二次调用事件...)。我正在堆栈如何获取父节点java/xpath中每个子节点的位置？

假设我有以下 xml:

<callEvents>
          <gprsCall>...</gprsCall>
          <gprsCall>...</gprsCall>
          <mobileOrigintaedCall>...</mobileOrigintaedCall>
          <gprsCall>...</gprsCall>
          <gprsCall>...</gprsCall>
</callEvents> 

所以代码应该返回:

• 我的第一个 gprsCall 位置 = 1
• 我的第二个 gprsCall 位置 = 2
• 我的第一个 mobileOriginatedCall 位置 = 3
• 我的第三个 gprsCall 位置 = 4
• 我的第四个 gprsCall 位置 = 5

Answer 1

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.xpath.XPathFactory;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathConstants;
import java.io.StringReader;
import org.xml.sax.InputSource;

class myxml {
    static final String xml = "<callEvents>\n"
        + "<gprsCall>...</gprsCall>\n"
        + "<gprsCall>...</gprsCall>\n"
        + "<mobileOrigintaedCall>...</mobileOrigintaedCall>\n"
        + "<gprsCall>...</gprsCall>\n"
        + "<gprsCall>...</gprsCall>\n"
        + "</callEvents>";

    public static void main (String... args) throws Throwable {
        try {
            DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
            Document doc = builder.parse(new InputSource(new StringReader(xml)));

            XPathExpression xpath = XPathFactory.newInstance().newXPath().compile("/callEvents/*");
            NodeList nodelist = (NodeList) xpath.evaluate(doc, XPathConstants.NODESET);
            for (int i = 0; i < nodelist.getLength(); i++) {
                System.out.println("Position: " + (i + 1)
                                   + ", name: " + nodelist.item(i).getNodeName());
            }
        } catch (Throwable e) {
            throw e;
        }
    }
}