我正在搜索可用於讀取XML文件的代碼。我確實找到了一個如下。但我的問題是,我無法在線閱讀XML文件。當我給出URL of the XML file位置時,它會返回「文件未找到異常」。可以有人建議。提前致謝。在線閱讀XML文件
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();
}
}
}
可能的重複[如何從Java中的URL讀取XML響應,](http://stackoverflow.com/questions/2310139/how-to-read-xml-response-from-a-url-in -java) – 2011-05-08 13:27:25