如何在ImageView中使用URL引用的图像?


当前回答

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

其他回答

你必须先下载图像

public static Bitmap loadBitmap(String url) {
    Bitmap bitmap = null;
    InputStream in = null;
    BufferedOutputStream out = null;

    try {
        in = new BufferedInputStream(new URL(url).openStream(), IO_BUFFER_SIZE);

        final ByteArrayOutputStream dataStream = new ByteArrayOutputStream();
        out = new BufferedOutputStream(dataStream, IO_BUFFER_SIZE);
        copy(in, out);
        out.flush();

        final byte[] data = dataStream.toByteArray();
        BitmapFactory.Options options = new BitmapFactory.Options();
        //options.inSampleSize = 1;

        bitmap = BitmapFactory.decodeByteArray(data, 0, data.length,options);
    } catch (IOException e) {
        Log.e(TAG, "Could not load Bitmap from: " + url);
    } finally {
        closeStream(in);
        closeStream(out);
    }

    return bitmap;
}

然后使用Imageview。setImageBitmap将位图设置为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。:)

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.

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());