[英]XML parse file from HTTP
我有一个XML文件位于如下的位置
http://example.com/test.xml
我正在尝试解析XML文件以在我的程序中使用它与xPath这样但它不起作用。
Document doc = builder.parse(new File(url));
我怎样才能获得XML文件?
尝试使用URLConnection.getInputStream()
来获取XML文件的句柄。
请参阅下面的代码,我试图打开一个xml文件并打印所有description
字段:
import java.io.InputStream;
import java.net.URL;
import java.net.URLConnection;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
public class HTTPXMLTest
{
public static void main(String[] args)
{
try {
new HTTPXMLTest().start();
} catch (Exception e) {
e.printStackTrace();
}
}
private void start() throws Exception
{
URL url = new URL("http://localhost:8080/AutoLogin/resource/web.xml");
URLConnection connection = url.openConnection();
Document doc = parseXML(connection.getInputStream());
NodeList descNodes = doc.getElementsByTagName("description");
for(int i=0; i<descNodes.getLength();i++)
{
System.out.println(descNodes.item(i).getTextContent());
}
}
private Document parseXML(InputStream stream)
throws Exception
{
DocumentBuilderFactory objDocumentBuilderFactory = null;
DocumentBuilder objDocumentBuilder = null;
Document doc = null;
try
{
objDocumentBuilderFactory = DocumentBuilderFactory.newInstance();
objDocumentBuilder = objDocumentBuilderFactory.newDocumentBuilder();
doc = objDocumentBuilder.parse(stream);
}
catch(Exception ex)
{
throw ex;
}
return doc;
}
}
以下是从此字符串“ http://www.gettingagile.com/feed/rss2/ ”获取数据的简单示例
public class MainClassXml {
public static void main(String args[]) throws URISyntaxException,
ClientProtocolException, IOException, MalformedURLException {
String url = "http://www.gettingagile.com/feed/rss2/";
System.out.println("Url is careated****");
URL url2 = new URL(url);
HttpGet httpGet = new HttpGet(url);
HttpClient httpClient = new DefaultHttpClient();
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity entity = httpResponse.getEntity();
System.out.println("Entity is*****" + entity);
try {
String xmlParseString = EntityUtils.toString(entity);
System.out.println("This Stirng to be Pasrse***" + xmlParseString);
HttpURLConnection connection = (HttpURLConnection) url2
.openConnection();
InputStream inputStream = connection.getInputStream();
DocumentBuilderFactory builderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder documentBuilder = builderFactory
.newDocumentBuilder();
Document document = documentBuilder.parse(inputStream);
document.getDocumentElement().normalize();
System.out.println("Attributes are***" + document.getAttributes());
NodeList nodeList = document.getElementsByTagName("rss");
System.out.println("This is firstnode" + nodeList);
for (int getChild = 0; getChild < nodeList.getLength(); getChild++) {
Node Listnode = nodeList.item(getChild);
System.out.println("Into the for loop"
+ Listnode.getAttributes().getLength());
Element firstnoderss = (Element) Listnode;
System.out.println("ListNodes" + Listnode.getAttributes());
System.out.println("This is node list length"
+ nodeList.getLength());
Node Subnode = nodeList.item(getChild);
System.out.println("This is list node" + Subnode);
System.out.println("rss attributes***************");
}
} catch (Exception exception) {
System.out.println("Exception is" + exception);
}
}
摆脱new File()
:
Document doc = builder.parse(url);
更多细节,基于laz答案:
String urlString = "http://example.com/test.xml";
URL url = new URL(urlString);
Document doc = builder.parse(url);
使用XMLPullParser会更容易......你不必处理这个事件,并且可以快速获取一些关键字......我也在使用它......只有几行代码:)
http://developer.android.com/reference/org/xmlpull/v1/XmlPullParser.html
关于HTTP和文件看看这里下载一个文件与DefaultHTTPClient和抢先认证
File fileXml = new File(url);
DocumentBuilder parser = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document document = parser.parse(fileXml);
它应该去
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