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

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" />

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.

要做到这一点,一个简单而干净的方法是使用开源库Prime。

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

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;
}