Getting Image from URL (Java)

Question

I am trying to read the following image

But it is showing IIOException.

Here is the code:

Image image = null;
URL url = new URL("http://bks6.books.google.ca/books?id=5VTBuvfZDyoC&printsec=frontcover&img=1& zoom=5&edge=curl&source=gbs_api");
image = ImageIO.read(url);
jXImageView1.setImage(image); 

Answer 1

This code worked fine for me.

 import java.io.FileOutputStream;
 import java.io.IOException;
 import java.io.InputStream;
 import java.io.OutputStream;
 import java.net.URL;

public class SaveImageFromUrl {

public static void main(String[] args) throws Exception {
    String imageUrl = "http://www.avajava.com/images/avajavalogo.jpg";
    String destinationFile = "image.jpg";

    saveImage(imageUrl, destinationFile);
}

public static void saveImage(String imageUrl, String destinationFile) throws IOException {
    URL url = new URL(imageUrl);
    InputStream is = url.openStream();
    OutputStream os = new FileOutputStream(destinationFile);

    byte[] b = new byte[2048];
    int length;

    while ((length = is.read(b)) != -1) {
        os.write(b, 0, length);
    }

    is.close();
    os.close();
}

}

Answer 2

You are getting an HTTP 400 (Bad Request) error because there is a space in your URL. If you fix it (before the zoom parameter), you will get an HTTP 401 error (Unauthorized). Maybe you need some HTTP header to identify your download as a recognised browser (use the "User-Agent" header) or additional authentication parameter.

For the User-Agent example, then use the ImageIO.read(InputStream) using the connection inputstream:

URLConnection connection = url.openConnection();
connection.setRequestProperty("User-Agent", "xxxxxx");

Use whatever needed for xxxxxx

Answer 3

Try This:

//urlPath = address of your picture on internet
URL url = new URL("urlPath");
BufferedImage c = ImageIO.read(url);
ImageIcon image = new ImageIcon(c);
jXImageView1.setImage(image);