Question

我正在搜索可以用来读取XML文件的代码。我确实找到了一个如下。但我的问题是，我无法在线阅读XML文件。当我给出URL of the XML file位置时，它返回File Not Found Exception。有人可以建议。提前致谢。

import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

public class XMLReader {

 public static void main(String argv[]) {

  try {
  File file = new File("MyXML.xml");
  DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
  DocumentBuilder db = dbf.newDocumentBuilder();
  Document doc = db.parse(file);
  doc.getDocumentElement().normalize();
  System.out.println("Root element " + doc.getDocumentElement().getNodeName());
  NodeList nodeLst = doc.getElementsByTagName("employee");
  System.out.println("Information of all employees");

  for (int s = 0; s < nodeLst.getLength(); s++) {

    Node fstNode = nodeLst.item(s);

    if (fstNode.getNodeType() == Node.ELEMENT_NODE) {

           Element fstElmnt = (Element) fstNode;
      NodeList fstNmElmntLst = fstElmnt.getElementsByTagName("firstname");
      Element fstNmElmnt = (Element) fstNmElmntLst.item(0);
      NodeList fstNm = fstNmElmnt.getChildNodes();
      System.out.println("First Name : "  + ((Node) fstNm.item(0)).getNodeValue());
      NodeList lstNmElmntLst = fstElmnt.getElementsByTagName("lastname");
      Element lstNmElmnt = (Element) lstNmElmntLst.item(0);
      NodeList lstNm = lstNmElmnt.getChildNodes();
      System.out.println("Last Name : " + ((Node) lstNm.item(0)).getNodeValue());
    }

  }
  } catch (Exception e) {
    e.printStackTrace();
  }
 }
}

Answer 1

它在stackoverflow上被禁止：How to read XML response from a URL in java?

Answer 2

您可以利用java.net.URL类：

URL xmlURL = new URL("http://www.cse.lk/listedcompanies/overview.htm?d-16544-e=3&6578706f7274=1");
InputStream xml = xmlURL.openStream();
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(xml);
xml.close();

Answer 3

如果您尝试与Restful Service进行通信，则可能会因使用库而受益。在这方面有好处的开源库包括Apache CXF和Jersey。

在线阅读XML文件

3 个答案: