在NetworkOnMainThreadException中将url转换为位图

Question

我想将某些URL中的图像动态添加到线性布局中.运行该段代码时出现错误无法启动活动

ComponentInfo:位于com.example的com.example.star.example.CompampleActivity.insertPhoto(CompinfoActivity.java:50)的com.example.star.example.CompinfoActivity.getBitmapFromURL(CompinfoActivity.java:70)中的android.os.NetworkOnMainThreadException. star.example.CompinfoActivity.onCreate(CompinfoActivity.java:38)

请帮忙.这是我下面活动的代码.

public class CompinfoActivity extends AppCompatActivity {

Dialog dialog;
LinearLayout myGallery;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_compinfo);

    dialog = new Dialog(this);
    dialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
    dialog.setContentView(R.layout.activity_compinfopopup);
    dialog.getWindow().getAttributes().width = AbsListView.LayoutParams.MATCH_PARENT;
    dialog.getWindow().getAttributes().height = AbsListView.LayoutParams.WRAP_CONTENT;
    dialog.show();

    myGallery = (LinearLayout) dialog.findViewById(R.id.mygallery);
    myGallery.addView(insertPhoto("http://example.com/a.png"));
    myGallery.addView(insertPhoto("http://example.com/b.png"));
    myGallery.addView(insertPhoto("http://example.com/c.png"));
    myGallery.addView(insertPhoto("http://example.com/d.png"));
    new MyTask().execute();



}

View insertPhoto(String path){
    Bitmap bm;
    bm =  getBitmapFromURL(path);
    LinearLayout layout = new LinearLayout(getApplicationContext());
    layout.setLayoutParams(new AbsListView.LayoutParams(250, 250));
    layout.setGravity(Gravity.CENTER);

    ImageView imageView = new ImageView(getApplicationContext());
    imageView.setLayoutParams(new AbsListView.LayoutParams(220, 220));
    imageView.setScaleType(ImageView.ScaleType.CENTER_CROP);
    imageView.setImageBitmap(bm);

    layout.addView(imageView);
    return layout;
}


public static Bitmap getBitmapFromURL(String src) {
    try {
        URL url = new 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;
    }
}

Answer 1

只需按以下方式操作 -

public class MyAsync extends AsyncTask<Void, Void, Bitmap>{

    @Override
    protected Bitmap doInBackground(Void... params) {

        try {
            URL url = new 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;
        }

    }
}

现在从url访问位图,如下所示 -

MyAsync obj = new MyAsync(){

        @Override
        protected void onPostExecute(Bitmap bmp) {
            super.onPostExecute(bmp);

            Bitmap bm = bmp;
            LinearLayout layout = new LinearLayout(getApplicationContext());
            layout.setLayoutParams(new AbsListView.LayoutParams(250, 250));
            layout.setGravity(Gravity.CENTER);

            ImageView imageView = new ImageView(getApplicationContext());
            imageView.setLayoutParams(new AbsListView.LayoutParams(220, 220));
            imageView.setScaleType(ImageView.ScaleType.CENTER_CROP);
            imageView.setImageBitmap(bm);

            layout.addView(imageView);
        }
    };

然后最终执行AsynTask -

obj.execute();