1. Picasso允许在应用程序中轻松加载图像——通常只需一行代码!

使用它:

implementation 'com.squareup.picasso:picasso:(insert latest version)'

只有一行代码!

Picasso.get().load("http://i.imgur.com/DvpvklR.png").into(imageView);

2. 一个用于Android的图像加载和缓存库，专注于平滑滚动

使用它:

repositories {
   mavenCentral() 
   google()
}

dependencies {
   implementation 'com.github.bumptech.glide:glide:4.11.0'
   annotationProcessor 'com.github.bumptech.glide:compiler:4.11.0'
}

// For a simple view:
Glide.with(this).load("http://i.imgur.com/DvpvklR.png").into(imageView);

3.fresco是一个在Android应用程序上显示图像的强大系统。Fresco负责图像加载和显示，所以你不必这样做。

从壁画开始

    String img_url= //url of the image
    URL url=new URL(img_url);
    Bitmap bmp; 
    bmp=BitmapFactory.decodeStream(url.openConnection().getInputStream());
    ImageView iv=(ImageView)findviewById(R.id.imageview);
    iv.setImageBitmap(bmp);

带有异常处理和异步任务的版本:

AsyncTask<URL, Void, Boolean> asyncTask = new AsyncTask<URL, Void, Boolean>() {
    public Bitmap mIcon_val;
    public IOException error;

    @Override
    protected Boolean doInBackground(URL... params) {
        try {
            mIcon_val = BitmapFactory.decodeStream(params[0].openConnection().getInputStream());
        } catch (IOException e) {
            this.error = e;
            return false;
        }
        return true;
    }

    @Override
    protected void onPostExecute(Boolean success) {
        super.onPostExecute(success);
        if (success) {
            image.setImageBitmap(mIcon_val);
        } else {
            image.setImageBitmap(defaultImage);
        }
    }
};
try {
    URL url = new URL(url);
    asyncTask.execute(url);
} catch (MalformedURLException e) {
    e.printStackTrace();
}

在我看来，最适合完成这项任务的现代图书馆是毕加索广场图书馆。它允许通过URL加载图像到ImageView，只需一行代码:

Picasso.with(context).load("http://i.imgur.com/DvpvklR.png").into(imageView);

这是一个迟到的回复，正如上面所建议的AsyncTask将会和谷歌一点后，我发现了一个解决这个问题的方法。

Drawable Drawable = Drawable. createfromstream ((InputStream) new URL(" URL ").getContent()， "src");

imageView.setImageDrawable(可拉的);

这是完整的功能:

public void loadMapPreview () {
    //start a background thread for networking
    new Thread(new Runnable() {
        public void run(){
            try {
                //download the drawable
                final Drawable drawable = Drawable.createFromStream((InputStream) new URL("url").getContent(), "src");
                //edit the view in the UI thread
                imageView.post(new Runnable() {
                    public void run() {
                        imageView.setImageDrawable(drawable);
                    }
                });
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }).start();
}

不要忘记在你的AndroidManifest.xml中添加以下权限来访问互联网。

< uses-permission android: name = " android.permission。互联网" / >

我自己也试过，还没有遇到任何问题。

public class LoadWebImg extends Activity {

String image_URL=
 "http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png";

   /** Called when the activity is first created. */
   @Override
   public void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.main);

       ImageView bmImage = (ImageView)findViewById(R.id.image);
    BitmapFactory.Options bmOptions;
    bmOptions = new BitmapFactory.Options();
    bmOptions.inSampleSize = 1;
    Bitmap bm = LoadImage(image_URL, bmOptions);
    bmImage.setImageBitmap(bm);
   }

   private Bitmap LoadImage(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 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;
}
}