如何在ImageView中使用URL引用的图像?


当前回答

imageView.setImageBitmap(BitmapFactory.decodeStream(imageUrl.openStream()));//try/catch IOException and MalformedURLException outside

其他回答

带有异常处理和异步任务的版本:

AsyncTask<URL, Void, Boolean> asyncTask = new AsyncTask<URL, Void, Boolean>() {
    public Bitmap mIcon_val;
    public IOException error;

    @Override
    protected Boolean doInBackground(URL... params) {
        try {
            mIcon_val = BitmapFactory.decodeStream(params[0].openConnection().getInputStream());
        } catch (IOException e) {
            this.error = e;
            return false;
        }
        return true;
    }

    @Override
    protected void onPostExecute(Boolean success) {
        super.onPostExecute(success);
        if (success) {
            image.setImageBitmap(mIcon_val);
        } else {
            image.setImageBitmap(defaultImage);
        }
    }
};
try {
    URL url = new URL(url);
    asyncTask.execute(url);
} catch (MalformedURLException e) {
    e.printStackTrace();
}

适用于任何容器中的imageView,比如listview grid view, normal layout

 private class LoadImagefromUrl extends AsyncTask< Object, Void, Bitmap > {
        ImageView ivPreview = null;

        @Override
        protected Bitmap doInBackground( Object... params ) {
            this.ivPreview = (ImageView) params[0];
            String url = (String) params[1];
            System.out.println(url);
            return loadBitmap( url );
        }

        @Override
        protected void onPostExecute( Bitmap result ) {
            super.onPostExecute( result );
            ivPreview.setImageBitmap( result );
        }
    }

    public Bitmap loadBitmap( String url ) {
        URL newurl = null;
        Bitmap bitmap = null;
        try {
            newurl = new URL( url );
            bitmap = BitmapFactory.decodeStream( newurl.openConnection( ).getInputStream( ) );
        } catch ( MalformedURLException e ) {
            e.printStackTrace( );
        } catch ( IOException e ) {

            e.printStackTrace( );
        }
        return bitmap;
    }
/** Usage **/
  new LoadImagefromUrl( ).execute( imageView, url );

要做到这一点,一个简单而干净的方法是使用开源库Prime。

如果你是在点击按钮的基础上加载图像,上面接受的答案是很棒的,但是如果你是在一个新的活动中做这件事,它会冻结UI一到两秒钟。环顾四周,我发现一个简单的asynctask消除了这个问题。

要使用asynctask,在activity的末尾添加这个类:

private class DownloadImageTask extends AsyncTask<String, Void, Bitmap> {
    ImageView bmImage;

    public DownloadImageTask(ImageView bmImage) {
        this.bmImage = bmImage;
    }

    protected Bitmap doInBackground(String... urls) {
        String urldisplay = urls[0];
        Bitmap mIcon11 = null;
        try {
            InputStream in = new java.net.URL(urldisplay).openStream();
            mIcon11 = BitmapFactory.decodeStream(in);
        } catch (Exception e) {
            Log.e("Error", e.getMessage());
            e.printStackTrace();
        }
        return mIcon11;
    }

    protected void onPostExecute(Bitmap result) {
        bmImage.setImageBitmap(result);
    }    
}

从你的onCreate()方法调用使用:

new DownloadImageTask((ImageView) findViewById(R.id.imageView1))
        .execute(MY_URL_STRING);

结果是一个快速加载的活动和一个稍后根据用户的网络速度显示的imageview。

我写了一个类来处理这个问题,因为它似乎是我各种项目中反复出现的需求:

https://github.com/koush/UrlImageViewHelper

UrlImageViewHelper will fill an ImageView with an image that is found at a URL. The sample will do a Google Image Search and load/show the results asynchronously. UrlImageViewHelper will automatically download, save, and cache all the image urls the BitmapDrawables. Duplicate urls will not be loaded into memory twice. Bitmap memory is managed by using a weak reference hash table, so as soon as the image is no longer used by you, it will be garbage collected automatically.