如何在ImageView中使用URL引用的图像?


当前回答

你也可以使用LoadingImageView视图从url加载图片:

http://blog.blundellapps.com/imageview-with-loading-spinner/

一旦你从链接中添加了类文件,你就可以实例化一个url图像视图:

在xml:

<com.blundell.tut.LoaderImageView
  android:id="@+id/loaderImageView"
  android:layout_width="wrap_content"
  android:layout_height="wrap_content"
  image="http://developer.android.com/images/dialog_buttons.png"
 />

在代码:

final LoaderImageView image = new LoaderImageView(this, "http://developer.android.com/images/dialog_buttons.png");

并更新它使用:

image.setImageDrawable("http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png");

其他回答

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.

    private Bitmap getImageBitmap(String url) {
        Bitmap bm = null;
        try {
            URL aURL = new URL(url);
            URLConnection conn = aURL.openConnection();
            conn.connect();
            InputStream is = conn.getInputStream();
            BufferedInputStream bis = new BufferedInputStream(is);
            bm = BitmapFactory.decodeStream(bis);
            bis.close();
            is.close();
       } catch (IOException e) {
           Log.e(TAG, "Error getting bitmap", e);
       }
       return bm;
    } 

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。

这段代码经过测试,它是完全工作的。

URL req = new URL(
"http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png"
);
Bitmap mIcon_val = BitmapFactory.decodeStream(req.openConnection()
                  .getInputStream());
public class LoadWebImg extends Activity {

String image_URL=
 "http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png";

   /** Called when the activity is first created. */
   @Override
   public void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.main);

       ImageView bmImage = (ImageView)findViewById(R.id.image);
    BitmapFactory.Options bmOptions;
    bmOptions = new BitmapFactory.Options();
    bmOptions.inSampleSize = 1;
    Bitmap bm = LoadImage(image_URL, bmOptions);
    bmImage.setImageBitmap(bm);
   }

   private Bitmap LoadImage(String URL, BitmapFactory.Options options)
   {       
    Bitmap bitmap = null;
    InputStream in = null;       
       try {
           in = OpenHttpConnection(URL);
           bitmap = BitmapFactory.decodeStream(in, null, options);
           in.close();
       } catch (IOException e1) {
       }
       return bitmap;               
   }

private InputStream OpenHttpConnection(String strURL) throws IOException{
 InputStream inputStream = null;
 URL url = new URL(strURL);
 URLConnection conn = url.openConnection();

 try{
  HttpURLConnection httpConn = (HttpURLConnection)conn;
  httpConn.setRequestMethod("GET");
  httpConn.connect();

  if (httpConn.getResponseCode() == HttpURLConnection.HTTP_OK) {
   inputStream = httpConn.getInputStream();
  }
 }
 catch (Exception ex)
 {
 }
 return inputStream;
}
}