Question

我试图在我的应用中显示来自网址的图片。但是我使用的方式很长。这个代码我建立在stackoverflow上

public  Bitmap getImage(String url,String src_name) throws java.net.MalformedURLException, java.io.IOException {
            Bitmap bitmap;
            HttpURLConnection connection = (HttpURLConnection)new URL(url) .openConnection();
            connection.setRequestProperty("User-agent","Mozilla/4.0");

            connection.connect();
            InputStream input= connection.getInputStream();

            bitmap = BitmapFactory.decodeStream(input);

            return bitmap;
}

在10-12秒内加载10张图像。如果使用此代码。

和

 ///==========================================================================================================================================
     public   Drawable getImage(String url, String src_name) throws java.net.MalformedURLException, java.io.IOException 
        {

         Drawable abc =Drawable.createFromStream(((java.io.InputStream)new java.net.URL(url).getContent()), src_name);

            return abc;


        }

如果使用此代码 - 图像在9-11秒内加载。图像不大。最大宽度或高度为400-450。我在循环中告诉这个函数是这样的：for (int i =0;i<10;i++){image[i]=getImage(url);} 可以告诉我如何在我的应用程序中最好地和紧固显示图像吗？亲爱的，彼得。

Answer 1

您无法取消下载和解码图像所需的时间。数字“10”只是图像质量的函数，您只能尝试优化此数字。

如果服务器由您管理，您可能需要花一些时间根据UI要求优化可下载图像的大小。也尝试延迟加载（我希望你不在UI线程上执行这些操作）。许多次讨论了懒惰下载和延迟解码的许多解决方案：http://www.google.com.sg/search?q=android+images+lazy+load&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

旁注：不鼓励使用HttpURLConnection。使用HttpClient。这也可能会影响性能。看看http://lukencode.com/2010/04/27/calling-web-services-in-android-using-httpclient/

Answer 2

public static Bitmap getBitmapFromUrl(String url) {
    Bitmap bitmap = null;
    HttpGet httpRequest = null;
    httpRequest = new HttpGet(url);
    HttpClient httpclient = new DefaultHttpClient();

    HttpResponse response = null;
    try {
        response = (HttpResponse) httpclient.execute(httpRequest);
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    if (response != null) {
        HttpEntity entity = response.getEntity();
        BufferedHttpEntity bufHttpEntity = null;
        try {
            bufHttpEntity = new BufferedHttpEntity(entity);
        } catch (IOException e) {
            e.printStackTrace();
        }

        InputStream instream = null;
        try {
            instream = bufHttpEntity.getContent();
        } catch (IOException e) {
            e.printStackTrace();
        }

        bitmap = BitmapFactory.decodeStream(instream);
    }
    return bitmap;
}

Answer 3

public static Bitmap decodeFile(String filePath) {
    // Decode image size
    BitmapFactory.Options o = new BitmapFactory.Options();
    o.inJustDecodeBounds = true;
    BitmapFactory.decodeFile(filePath, o);

    // The new size we want to scale to
    final int REQUIRED_SIZE = 1024;

    // Find the correct scale value. It should be the power of 2.
    int width_tmp = o.outWidth, height_tmp = o.outHeight;
    int scale = 1;
    while (true) {
        if (width_tmp < REQUIRED_SIZE && height_tmp < REQUIRED_SIZE)
            break;
        width_tmp /= 2;
        height_tmp /= 2;
        scale *= 2;
    }

    // Decode with inSampleSize
    BitmapFactory.Options o2 = new BitmapFactory.Options();
    o2.inSampleSize = scale;
    Bitmap bitmap = BitmapFactory.decodeFile(filePath, o2);
    return bitmap;
}

从url到drawable或bitmap的图像：最好和最快的方式

3 个答案: