由于URL中的空格而无法下载图像

Question

I have an application where i download images from a server and display them in a imageview. This works the majority of the time but not always. It seems the times it doesn't work is when there are spaces in the url. The error i get is

java.lang.RuntimeException: java.io.FileNotFoundException:

I have tried a number of different ways to try and encode the url but have had no success upto now.

Here is the class I am using.

private URL url;
Bitmap bitmap;

public ImageDownloader() {

}//constructor


public Drawable getImage(String urlString) throws IOException{
    Log.i("url", urlString);
    url = new URL(urlString);
    InputStream is = url.openStream();
    Bitmap bitmap = BitmapFactory.decodeStream(new FlushedInputStream(is));
    Drawable image = new BitmapDrawable(bitmap);
    return image;

 }//getImage

static class FlushedInputStream extends FilterInputStream {
    public FlushedInputStream(InputStream inputStream) {
        super(inputStream);
    }

    @Override
    public long skip(long n) throws IOException {
        long totalBytesSkipped = 0L;
        while (totalBytesSkipped < n) {
            long bytesSkipped = in.skip(n - totalBytesSkipped);
            if (bytesSkipped == 0L) {
                  int bytee = read();
                  if (bytee < 0) {
                      break;  // we reached EOF
                  } else {
                      bytesSkipped = 1; // we read one byte
                  }
           }
            totalBytesSkipped += bytesSkipped;
        }
        return totalBytesSkipped;
    }
}


public Bitmap getBitmap(String urlString) {
    try {
        //String s = Uri.encode(urlString);
        url = new URL(urlString);
    } catch (MalformedURLException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
    HttpURLConnection conn = null;
    try {
        conn = (HttpURLConnection) url.openConnection();
           conn.setRequestProperty("User-agent", "Mozilla/4.0");
           conn.connect();

    } catch (IOException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }

    InputStream in = null;
    try {
        in = conn.getInputStream();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return bitmap = BitmapFactory.decodeStream(in);



}

Both the getImage and getBitmap have the same error.

Here is an example of a url i am using.

http://kaiorize-clone1.loomarea.com/sites/default/files/imagecache/app_product_full/sites/default/files/BsfL_Logo_final_sonderfarben 300dpi.jpg

Answer 1

This was a strange one. I tried to encode the normal way but I was still getting the same error. I check the URL that was being sent and the space had been changed for a "%20". In the end through desperation I decided to just change the space into "%20" myself instead of encoding. This solved the problem. Not the most elegent of solutions but it worked!

        String url1 = json.getString("app_imagepath");
        String url = url1.replace(" ", "%20");

Answer 2

URL encoding converts characters into a format that can be transmitted over the Internet. For example, space is encoded as %20.

You can find the entire table at http://www.w3schools.com/tags/ref_urlencode.asp .

Answer 3

Try this code snippet,

String songThumbUrl = "Image URL";
        try {
        URL thumbURL = new URL(songThumbUrl);
        try {
            URLConnection urlConnection = thumbURL.openConnection();
            urlConnection.connect();

            InputStream inputStream = urlConnection.getInputStream();
            BufferedInputStream bufferedInputStream = new BufferedInputStream(inputStream);
            Bitmap bitmap = BitmapFactory.decodeStream(bufferedInputStream);
            songThumbImageView.setImageBitmap(bitmap);

            bufferedInputStream.close();
            inputStream.close();

        } catch (IOException ioe) {
            System.out.println("My Exception :" + ioe);
            ioe.printStackTrace();
        }
    } catch (MalformedURLException mue) {
        System.out.println("My Exception :" + mue);
        mue.printStackTrace();
    }