java 在线读取 XML 文件
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Reading XML file online
提问by Jay Mayu
I was searching code that I can use to read XML file. and I did find one as below. But my problem is, I'm unable to read a XML file online. When I give the URL of the XML filelocation, it returns File Not Found Exception. Can someone advice. Thanks in advance.
我正在搜索可用于读取 XML 文件的代码。我确实找到了一个如下。但我的问题是，我无法在线读取 XML 文件。当我提供XML 文件位置的URL 时，它返回文件未找到异常。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();
}
}
}
回答by Marcin
It was discoused on stackoverflow: How to read XML response from a URL in java?
它在 stackoverflow: How to read XML response from a URL in java?
回答by bdoughan
You can leverage the java.net.URL class:
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();
回答by bmargulies
If you are trying to communicate with a Restful Service, you might benefit from using a library. Open source libraries with goodies in this area include Apache CXF and Jersey.
如果您尝试与 Restful 服务通信，您可能会从使用库中受益。在这方面具有优势的开源库包括 Apache CXF 和 Jersey。
