尽管存在图像,但BitmapFactory返回null

时间:2015-12-16 07:21:04

标签: android bitmapfactory

这里我想从String URL转换图像。虽然有一个包含图像的URL,但它返回null。我在下面分享了代码。

private byte[] convertImageToByteArray(String imgPath)
{

    byte[] byteArray = null;
    Bitmap bmp = BitmapFactory.decodeFile(imgPath);
    if(bmp != null)
    {

        try {
            ByteArrayOutputStream stream = new ByteArrayOutputStream();
            //bmp.compress(Bitmap.CompressFormat.JPEG, 100, stream);
            bmp.compress(Bitmap.CompressFormat.PNG, 100, stream);
            byteArray = stream.toByteArray();

            try 
            {
                stream.close();
            } 
            catch (IOException e) 
            {
                e.printStackTrace();

            }
        } catch (Exception e) {
            e.printStackTrace();

        }
    }
    else
    {
        try {
            Bitmap bmpDefault = BitmapFactory.decodeResource(getResources(), R.drawable.na);
            ByteArrayOutputStream stream = new ByteArrayOutputStream();
            //bmpDefault.compress(Bitmap.CompressFormat.JPEG, 100, stream);
            bmpDefault.compress(Bitmap.CompressFormat.PNG, 100, stream);
            byteArray = stream.toByteArray();
        } 
        catch (Exception e) 
        {
            e.printStackTrace();

        }

    }
return byteArray;

}

控制流不是执行if块,而是进入else块,而BitmapFactory.decodeFile()总是返回null。哪里出错了?

3 个答案:

答案 0 :(得分:3)

Ravindra的答案很有帮助,为了更好地利用图像尝试Picasso lib,令人兴奋。它还调整了大小/裁剪方法。

Picasso.with(getContext()).load("your url").into(new Target() {
                    @Override
                    public void onBitmapLoaded(Bitmap bitmap, Picasso.LoadedFrom from) {
                        //do what ever you want with your bitmap 
                    }

                    @Override
                    public void onBitmapFailed(Drawable errorDrawable) {

                    }

                    @Override
                    public void onPrepareLoad(Drawable placeHolderDrawable) {

                    }
                });

答案 1 :(得分:2)

您可以使用此引用,它可能是helpfull

注意: - 此功能进行网络连接,您应该在线程或AsyncTask中调用它。否则可能会抛出code = textwrap.dedent("""\ val NUM_SAMPLES = 100000; val count = sc.parallelize(1 to NUM_SAMPLES).map { i => val x = Math.random(); val y = Math.random(); if (x*x + y*y < 1) 1 else 0 }.reduce(_ + _); println(\"Pi is roughly \" + 4.0 * count / NUM_SAMPLES) """) data = { 'code': code } r = requests.post(statements_url, data=json.dumps(data), headers=headers) 异常。

当函数返回Bitmap时,您必须等到线程执行完毕,请检查question。 它使用join()

我希望这会有所帮助。

答案 2 :(得分:1)

使用以下代码行: -

    Bitmap bmImg;
   AsyncTask<String, String, String> _Task = new AsyncTask<String, String, String>()
{
    @Override
    protected void onPreExecute()
    {

        String _imgURL  = "**HERE IS YOUR URL OF THE IMAGE**";

    }

    @Override
    protected String doInBackground(String... arg0)
    {


            HttpGet httpRequest = new HttpGet(URI.create(_imgURL));
            HttpClient httpclient = new DefaultHttpClient();
            HttpResponse response = (HttpResponse) httpclient.execute(httpRequest);
            HttpEntity entity = response.getEntity();
            BufferedHttpEntity bufHttpEntity = new BufferedHttpEntity(entity);
            bmImg = BitmapFactory.decodeStream(bufHttpEntity.getContent());
            System.out.println("main url"+mainUrl);
            httpRequest.abort();


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



        return null;

    }

    @Override
    protected void onPostExecute(String result)
    {
        try
        {
            /////HERE USE YOURS BITMAP
         **bmImg** is your bitmap , you can use it any where i your class
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }
};
_Task.executeOnExecutor(AsyncTask.THREAD_POOL_EXECUTOR);