Open URL in Java to get the content

Question

I´m searching for a opportunity to open a url in java.

URL url = new URL("http://maps.google.at/maps?saddr=4714&daddr=Marchtrenk&hl=de");
    InputStream is = url.openConnection().getInputStream();

    BufferedReader reader = new BufferedReader( new InputStreamReader( is )  );

    String line = null;
    while( ( line = reader.readLine() ) != null )  {
       System.out.println(line);
    }
    reader.close();

I found that way.

I added it in my program and the following error occurred.

The method openConnection() is undefined for the type URL

(by url.openConnection())

What is my problem?

I use a tomcat-server with servlets, ...

Answer 1

Following code should work,

URL url = new URL("http://maps.google.at/maps?saddr=4714&daddr=Marchtrenk&hl=de");
InputStream is = url.openConnection().getInputStream();

BufferedReader reader = new BufferedReader( new InputStreamReader( is )  );

String line = null;
while( ( line = reader.readLine() ) != null )  {
   System.out.println(line);
}
reader.close();

Answer 2

It may be more useful to use a http client library like such as this

There are more things like access denied , document moved etc to handle when dealing with http.

(though, it is unlikely in this case)

Answer 3

Are you sure using the java.net.URL class? Check your import statements.

Answer 4

If you just want to open up the webpage, I think less is more in this case:

import java.awt.Desktop;
import java.net.URI; //Note this is URI, not URL

class BrowseURL{
    public static void main(String args[]) throws Exception{
        // Create Desktop object
        Desktop d=Desktop.getDesktop();

        // Browse a URL, say google.com
        d.browse(new URI("http://google.com"));

        }
    }
}

Answer 5

public class UrlContent{
    public static void main(String[] args) {

        URL url;

        try {
            // get URL content

            String a="http://localhost:8080/TestWeb/index.jsp";
            url = new URL(a);
            URLConnection conn = url.openConnection();

            // open the stream and put it into BufferedReader
            BufferedReader br = new BufferedReader(
                               new InputStreamReader(conn.getInputStream()));

            String inputLine;
            while ((inputLine = br.readLine()) != null) {
                    System.out.println(inputLine);
            }
            br.close();

            System.out.println("Done");

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }
}

Answer 6

I found this question while Googling. Note that if you just want to make use of the URI's content via something like a string, consider using Apache's IOUtils.toString() method.

For example, a sample line of code could be:

String pageContent = IOUtils.toString("http://maps.google.at/maps?saddr=4714&daddr=Marchtrenk&hl=de", Charset.UTF_8);