Android:从网址加载图片

Question

可能重复:
加载远程图像

我正在开发一个应用程序,其中我有一个列表,其中每行都有一个图像,

此图像是从Web URL加载的.

下面是我的代码

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="match_parent"
    android:layout_height="match_parent" >

    <ImageView
            android:id="@+id/list_row_img"
            android:layout_width="200dp"
            android:layout_height="200dp"
            />

    <TextView
            android:id="@+id/list_row_description"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            />

</LinearLayout>

这是活动代码

@Override
    public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    ImageView img= (ImageView) findViewById(R.id.list_row_img);
    Bitmap bm = null;
    try {
         URL aURL = new URL("http://image10.bizrate-images.com/resize?sq=60&uid=2216744464");
         URLConnection conn = aURL.openConnection();
         conn.connect();
         InputStream is = conn.getInputStream();
         BufferedInputStream bis = new BufferedInputStream(is);
         bm = BitmapFactory.decodeStream(bis);
         img.setImageBitmap(bm);
         bis.close();
         is.close();

         } catch (Exception e) {
            Log.v("EXCEPTION", "Error getting bitmap", e);
         }
}

当我跑步时,我在设备上什么也得不到.只有灰色屏幕,没有异常发生.

请注意,我在清单文件中添加了此权限

<uses-permission android:name="android.permission.INTERNET" />

有人可以帮我吗？

Answer 1

试试这个你可以执行它

      new DownloadImageTask((ImageView)  im1).execute(url here);

    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;
    }


/*    @Override
    protected void onPreExecute() {
        // TODO Auto-generated method stub
        super.onPreExecute();
        showProgressDialog();
    }*/

    protected void onPostExecute(Bitmap result) {
        //pDlg.dismiss();
        bmImage.setImageBitmap(result);
    }}

Answer 2

使用以下代码从url下载图像并将图像显示到imageview,这将解决您的问题.

import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import android.os.Bundle;
import android.os.StrictMode;
import android.app.Activity;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.widget.ImageView;

public class MainActivity extends Activity {

    ImageView mImgView1;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder()
                .permitAll().build();
        StrictMode.setThreadPolicy(policy);

        mImgView1 = (ImageView) findViewById(R.id.mImgView1);
        String url = "https://www.morroccomethod.com/components/com_virtuemart/shop_image/category/resized/Trial_Sizes_4e4ac3b0d3491_175x175.jpg";
        BitmapFactory.Options bmOptions;
        bmOptions = new BitmapFactory.Options();
        bmOptions.inSampleSize = 1;
        Bitmap bm = loadBitmap(url, bmOptions);
        mImgView1.setImageBitmap(bm);
    }

    public static Bitmap loadBitmap(String URL, BitmapFactory.Options options) {
        Bitmap bitmap = null;
        InputStream in = null;
        try {
            in = OpenHttpConnection(URL);
            bitmap = BitmapFactory.decodeStream(in, null, options);
            in.close();
        } catch (IOException e1) {
        }
        return bitmap;
    }

    private static InputStream OpenHttpConnection(String strURL)
            throws IOException {
        InputStream inputStream = null;
        URL url = new URL(strURL);
        URLConnection conn = url.openConnection();

        try {
            HttpURLConnection httpConn = (HttpURLConnection) conn;
            httpConn.setRequestMethod("GET");
            httpConn.connect();

            if (httpConn.getResponseCode() == HttpURLConnection.HTTP_OK) {
                inputStream = httpConn.getInputStream();
            }
        } catch (Exception ex) {
        }
        return inputStream;
    }
}