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

嗨,我有最简单的代码试试这个

    public class ImageFromUrlExample extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.main);  
            ImageView imgView =(ImageView)findViewById(R.id.ImageView01);
            Drawable drawable = LoadImageFromWebOperations("http://www.androidpeople.com/wp-content/uploads/2010/03/android.png");
            imgView.setImageDrawable(drawable);

    }

    private Drawable LoadImageFromWebOperations(String url)
    {
          try{
        InputStream is = (InputStream) new URL(url).getContent();
        Drawable d = Drawable.createFromStream(is, "src name");
        return d;
      }catch (Exception e) {
        System.out.println("Exc="+e);
        return null;
      }
    }
   }

main。xml

  <LinearLayout 
    android:id="@+id/LinearLayout01"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    xmlns:android="http://schemas.android.com/apk/res/android">
   <ImageView 
       android:id="@+id/ImageView01"
       android:layout_height="wrap_content" 
       android:layout_width="wrap_content"/>

试试这个

要做到这一点,一个简单而干净的方法是使用开源库Prime。

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 );

这里有很多好的信息…我最近发现了一个叫SmartImageView的类,到目前为止它似乎工作得很好。非常容易合并和使用。

http://loopj.com/android-smart-image-view/

https://github.com/loopj/android-smart-image-view

更新:我最后写了一篇关于这个的博客文章,所以看看它对使用SmartImageView的帮助。

第二次更新:我现在总是用毕加索做这个(见上文),强烈推荐它。:)

    String img_url= //url of the image
    URL url=new URL(img_url);
    Bitmap bmp; 
    bmp=BitmapFactory.decodeStream(url.openConnection().getInputStream());
    ImageView iv=(ImageView)findviewById(R.id.imageview);
    iv.setImageBitmap(bmp);