如何在ImageView中使用URL引用的图像?


当前回答

来自Android开发者:

// show The Image in a ImageView
new DownloadImageTask((ImageView) findViewById(R.id.imageView1))
            .execute("http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png");

public void onClick(View v) {
    startActivity(new Intent(this, IndexActivity.class));
    finish();

}

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);
    }
}

确保你在AndroidManifest.xml中设置了以下权限来访问互联网。

<uses-permission android:name="android.permission.INTERNET" />

其他回答

Android Query可以为你处理这个问题,还有更多(比如缓存和加载进度)。

看看这里。

我认为这是最好的方法。

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 );

在我看来,最适合完成这项任务的现代图书馆是毕加索广场图书馆。它允许通过URL加载图像到ImageView,只需一行代码:

Picasso.with(context).load("http://i.imgur.com/DvpvklR.png").into(imageView);

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。

这将帮助你……

定义imageview并将图像加载到.....

Imageview i = (ImageView) vv.findViewById(R.id.img_country);
i.setImageBitmap(DownloadFullFromUrl(url));

然后定义这个方法:

    public Bitmap DownloadFullFromUrl(String imageFullURL) {
    Bitmap bm = null;
    try {
        URL url = new URL(imageFullURL);
        URLConnection ucon = url.openConnection();
        InputStream is = ucon.getInputStream();
        BufferedInputStream bis = new BufferedInputStream(is);
        ByteArrayBuffer baf = new ByteArrayBuffer(50);
        int current = 0;
        while ((current = bis.read()) != -1) {
            baf.append((byte) current);
        }
        bm = BitmapFactory.decodeByteArray(baf.toByteArray(), 0,
                baf.toByteArray().length);
    } catch (IOException e) {
        Log.d("ImageManager", "Error: " + e);
    }
    return bm;
}