Question

我在我的应用程序中使用json解析。我有ListView显示一些数据和ImageView。这里我想显示来自ImageView中另一个网址的图片。怎么做？

这是我的代码......

private ProgressDialog pDialog;
private static String url = "http://myurl/json/main.json";

// JSON Node names

private static final String TAG_NAME = "name";
private static final String TAG_VIEWERS = "viewers";

    JSONArray myarray = null;

    ArrayList<HashMap<String, String>> mylist;

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



    ImageView bmImage = (ImageView) findViewById(R.id.img);


    mylist = new ArrayList<HashMap<String, String>>();
    ListView lv = getListView();
            lv.setOnItemClickListener(new OnItemClickListener() {

        @Override
        public void onItemClick(AdapterView<?> parent, View view,
                int position, long id) {
                            String name = ((TextView) view.findViewById(R.id.name))
                    .getText().toString();


        }
    });

            new GetMyarray().execute();


    }


/**
 * Async task class to get json by making HTTP call
 * */
private class GetMyarray extends AsyncTask<Void, Void, Void> {

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        // Showing progress dialog
        pDialog = new ProgressDialog(MainActivity.this);
        pDialog.setMessage("Processing...");
        pDialog.setCancelable(false);
        pDialog.show();

    }

    @Override
    protected Void doInBackground(Void... arg0) {
        // Creating service handler class instance
        ServiceHandler sh = new ServiceHandler();

        // Making a request to url and getting response
        String jsonStr = sh.makeServiceCall(url, ServiceHandler.GET);

        Log.d("Response: ", "> " + jsonStr);

        if (jsonStr != null) {
            try {

                JSONObject getResponse = new JSONObject(jsonStr);
                Log.d("Response: ", "> " + jsonStr);

                JSONObject pnames=getResponse.getJSONObject("getnames");
                Log.d("Response: ", "> " + pnames);



                                    myarray = pnames.getJSONArray("pnames");
                Log.d("Listed as: ", "> " + myarray);


                for (int i = 0; i < myarray.length(); i++) {
                    JSONObject c = myarray.getJSONObject(i);


                    String name = c.getString(TAG_NAME);
                    String viewers=c.getString(TAG_VIEWERS);



                    HashMap<String, String> myarray = new HashMap<String, String>();



                    myarray.put(TAG_NAME, name);
                    myarray.put(TAG_VIEWERS, viewers);




                    mylist.add(myarray);


                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        } else {
            Log.e("ServiceHandler", "Couldn't get any data from the url");
        }

        return null;
    }

    @Override
    protected void onPostExecute(Void result) {
        super.onPostExecute(result);
        // Dismiss the progress dialog
        if (pDialog.isShowing())
            pDialog.dismiss();

        /**
         * Updating parsed JSON data into ListView
         * */

        ListAdapter adapter = new SimpleAdapter(
                MainActivity.this, mylist,
                R.layout.list_item, new String[] { TAG_NAME,TAG_VIEWERS  }, new int[] { R.id.name,R.id.viewers
                        });

        setListAdapter(adapter);
    }

}

}

Answer 1

Here它展示了如何从URL

加载listview中的图像

只是为了从URL加载图像，有很多SO线程

Answer 2

请访问

用于延迟加载实现的listview

https://github.com/square/picasso

https://github.com/nostra13/Android-Universal-Image-Loader

https://github.com/thest1/LazyList

Answer 3

尝试这种方法

URL url = new URL("your url");
Bitmap bmp = BitmapFactory.decodeStream(url.openConnection().getInputStream());
imageView.setImageBitmap(bmp);

和

public boolean loadImageFromURL(String url, ImageView iv){
try {

URL myUrl = new URL (url);
HttpURLConnection conn =
  (HttpURLConnection) myUrl.openConnection();
conn.setDoInput(true);
conn.connect();

InputStream is = conn.getInputStream();
iv.setImageBitmap(BitmapFactory.decodeStream(is));

return true;

}  catch (Exception e) {
e.printStackTrace();
 }

return false;
}

* 记住：在Asyntask中执行这两个方法。否则会崩溃

如何在android中的ListView中显示来自url的图像？

3 个答案: