0

我正在尝试从我正在使用的 Web 服务设置图像：

private class FetchImageTask extends AsyncTask<String, Integer, Bitmap> {
    @Override
    protected Bitmap doInBackground(String... arg0) {
        Bitmap b = null;
        try {
            b = BitmapFactory.decodeStream((InputStream) new URL(arg0[0]).getContent());
        }
        catch (MalformedURLException e) {
            e.printStackTrace();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return b;
    }
}

并试图获取它

final ImageView imgicon = (ImageView) convertView.findViewById(R.id.imgicon); 
    new FetchImageTask() {
        @Override
        protected void onPostExecute(Bitmap result) {

            if (result != null) {
                imgicon.setImageBitmap(result);

            }
        }
    }.execute("Url/images/"+bitmapname);

但它不显示它也没有任何错误。有什么猜测吗？

4

5 回答 5

1

来自 Android 开发者博客

使用此异步任务

class BitmapDownloaderTask extends AsyncTask<String, Void, Bitmap> {
private String url;
private final WeakReference<ImageView> imageViewReference;

public BitmapDownloaderTask(ImageView imageView) {
    imageViewReference = new WeakReference<ImageView>(imageView);
}

@Override
// Actual download method, run in the task thread
protected Bitmap doInBackground(String... params) {
     // params comes from the execute() call: params[0] is the url.
     return downloadBitmap(params[0]);
}

@Override
// Once the image is downloaded, associates it to the imageView
protected void onPostExecute(Bitmap bitmap) {
    if (isCancelled()) {
        bitmap = null;
    }

    if (imageViewReference != null) {
        ImageView imageView = imageViewReference.get();
        if (imageView != null) {
            imageView.setImageBitmap(bitmap);
        }
    }
}
}

使用此功能从 url 下载位图

static Bitmap downloadBitmap(String url) {
final AndroidHttpClient client = AndroidHttpClient.newInstance("Android");
final HttpGet getRequest = new HttpGet(url);

try {
    HttpResponse response = client.execute(getRequest);
    final int statusCode = response.getStatusLine().getStatusCode();
    if (statusCode != HttpStatus.SC_OK) { 
        Log.w("ImageDownloader", "Error " + statusCode + " while retrieving bitmap from " + url); 
        return null;
    }

    final HttpEntity entity = response.getEntity();
    if (entity != null) {
        InputStream inputStream = null;
        try {
            inputStream = entity.getContent(); 
            final Bitmap bitmap = BitmapFactory.decodeStream(inputStream);
            return bitmap;
        } finally {
            if (inputStream != null) {
                inputStream.close();  
            }
            entity.consumeContent();
        }
    }
} catch (Exception e) {
    // Could provide a more explicit error message for IOException or IllegalStateException
    getRequest.abort();
    Log.w("ImageDownloader", "Error while retrieving bitmap from " + url, e.toString());
} finally {
    if (client != null) {
        client.close();
    }
}
return null;
}

如下调用 asynctask

ImageView mImageView = (ImageView)findViewById(yourImageViewId);
BitmapDownloaderTask mDownloaderTask = new BitmapDownloaderTask(mImageView);
mDownloaderTask.execute("YourDownloadUrlHere");
于 2013-10-29T06:44:18.853 回答
0

  1. 下载AndroidQuery罐子。

  2. 将此 jar 放入您的libs文件夹并右键单击 jar 和Build Path -> Add to build path

根据此示例使用它：

AQuery androidQuery = new AQuery(this); // make AndroidQuery object

androidQuery
    .id(yourImageView)
    .image(
        imageUrl,
        isCacheUrlImageOnMemery, 
        isCacheUrlImageOnFile);

如果为真，则在内存上给定 URL 图像缓存，因此在单词 android 查询检查后，在内存或文件上给出 URL 图像缓存，然后它从缓存中获取，否则它尝试从 URL 获取

isCacheUrlImageOnMemery

相同，isCacheUrlImageOnMemery但首先 android 查询检查内存然后文件，以防我们没有太多内存，然后我们缓存在文件中，这两个选项可用。

isCacheUrlImageOnFile
于 2013-10-29T06:44:19.903 回答
0

做这样的事情：

private class FetchImageTask extends AsyncTask<String, Integer, Bitmap> {
    @Override
    protected Bitmap doInBackground(String... arg0) {
        Bitmap b = null;
        try {
            b = BitmapFactory.decodeStream((InputStream) new URL(arg0[0]).getContent());
        }
        catch (MalformedURLException e) {
            e.printStackTrace();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return b;
    }
    @Override
    protected void onPostExecute(Bitmap result) {

        if (result != null) {
            imgicon.setImageBitmap(result);

        }
    }
}

然后这样称呼它：

final ImageView imgicon = (ImageView) convertView.findViewById(R.id.imgicon); 
new FetchImageTask().execute("Url/images/"+bitmapname);
于 2013-10-29T06:37:18.140 回答
0

尝试这个：

public Bitmap getBitmapFromURL(String src) {
    try {
        java.net.URL url = new java.net.URL(src);
        HttpURLConnection connection = (HttpURLConnection) url
                .openConnection();
        connection.setDoInput(true);
        connection.connect();
        InputStream input = connection.getInputStream();
        Bitmap myBitmap = BitmapFactory.decodeStream(input);
        return myBitmap;
    } catch (IOException e) {
        e.printStackTrace();
        return null;
    }
}

对于 OutOfMemoryIssue 使用：

public Bitmap getResizedBitmap(Bitmap bm, int newHeight, int newWidth) {
    int width = bm.getWidth();
    int height = bm.getHeight();
    float scaleWidth = ((float) newWidth) / width;
    float scaleHeight = ((float) newHeight) / height;
    // CREATE A MATRIX FOR THE MANIPULATION
    Matrix matrix = new Matrix();
    // RESIZE THE BIT MAP
    matrix.postScale(scaleWidth, scaleHeight);

    // "RECREATE" THE NEW BITMAP
    Bitmap resizedBitmap = Bitmap.createBitmap(bm, 0, 0, width, height,
            matrix, false);

    return resizedBitmap;
}
于 2013-10-29T06:59:13.533 回答
0 
private class FetchImageTask extends AsyncTask<String, Integer, Bitmap> {
    @Override
    protected void doInBackground(String... arg0) {
        try {
        java.net.URL url = new java.net.URL(arg0[0]);
        HttpURLConnection connection = (HttpURLConnection) url
                .openConnection();
        connection.setDoInput(true);
        connection.connect();
        InputStream input = connection.getInputStream();
        Bitmap myBitmap = BitmapFactory.decodeStream(input);

    } catch (IOException e) {
        e.printStackTrace();
        return null;
    }
        return null;
    }
}
@Override
    protected void onPostExecute(Bitmap result) {


            imageview.setImageBitmap(mybitmap);

        }
    }
于 2013-10-29T07:13:01.013 回答