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, ...

try this one: http://docs.oracle.com/javase/tutorial/networking/urls/readingURL.html — shem, Apr 02 '12 at 13:21
The problem is *probably* that you're not using `java.net.URL` but some other class called `URL`. — Joachim Sauer, Apr 02 '12 at 13:27

score 19 · Answer 1 · answered Feb 04 '13 at 12:18

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();
        }

    }
}

score 15 · Answer 2 · edited May 06 '13 at 08:14

15

String url_open ="http://javadl.sun.com/webapps/download/AutoDL?BundleId=76860";
java.awt.Desktop.getDesktop().browse(java.net.URI.create(url_open));

edited May 06 '13 at 08:14

Jens Erat

37,523
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answered May 06 '13 at 07:58

miguepiscy

151
1
2

works like a charm. thank you – Phil Apr 19 '22 at 19:24

score 6 · Answer 3 · edited Jul 01 '14 at 12:10

6

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();

edited Jul 01 '14 at 12:10

Kishan_KP

4,488
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answered Jul 01 '14 at 11:43

deepak bhat

61
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1

score 6 · Answer 4 · answered Apr 02 '12 at 13:29

6

It works for me. Please check if you are using the right imports?

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.URL;

answered Apr 02 '12 at 13:29

Boosty

701
1
5
12

score 2 · Answer 5 · answered Jan 02 '16 at 07:30

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"));

        }
    }
}

score 2 · Answer 6 · answered Apr 02 '12 at 13:29

2

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

answered Apr 02 '12 at 13:29

rlegendi

10,466
3
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score 1 · Answer 7 · answered Jan 13 '19 at 21:47

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);

score 1 · Answer 8 · answered Apr 02 '12 at 13:30

1

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)

answered Apr 02 '12 at 13:30

Jayan

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Open URL in Java to get the content

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