Question

我正在尝试调用一个动态生成按钮的方法onPostExecute方法的AsyncTask.OnPostExecute结果可以像S~D~W~H~H~A这样的结果。我需要动态创建按钮

private TextView textView;
        LinearLayout layout;
        Button btn;
        List<String> elements;  
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main);
            textView = (TextView) findViewById(R.id.TextView01);
             layout= (LinearLayout) findViewById(R.id.layoutFor);

        }

        private class ExecuteTask extends AsyncTask<String, Void, String> {
          @Override

            protected String doInBackground(String... urls) {
              String response = "";
              for (String url : urls) {
                DefaultHttpClient client = new DefaultHttpClient();
                HttpGet httpGet = new HttpGet(url);
                try {
                  HttpResponse execute = client.execute(httpGet);
                  InputStream content = execute.getEntity().getContent();
                  BufferedReader buffer = new BufferedReader(new InputStreamReader(content));
                  String s = "";
                  while ((s = buffer.readLine()) != null) {
                    response += s;
                  }          
                } catch (Exception e) {
                  e.printStackTrace();
                }
              }
              return response;
            }
            @Override
            protected void onPostExecute(String result) {

          elements = Arrays.asList(result.split("~"));  

                 generateButtonDynamically();
            }
          }
          public void onClick(View view) {
                    ExecuteTask task = new ExecuteTask ();
               task.execute(new String[] {"url"});
           }



      public void generateButtonDynamically()
        {

   for (int i = 0; i < elements.size(); i++) {

  LinearLayout linearLayout = (LinearLayout)findViewById(R.id.layoutFor);
                    Button btn = new Button(this); 
                    btn.setText(elements.get(i)); 
                    linearLayout.addView(btn);  
            }



        }

在generateButtonDynamically方法中。我正在尝试动态生成按钮并显示。但它没有发生。

Answer 1

不需要在 onPostExecute（） 方法中再次找到 LinearLayout 的ID。你已经全球宣布了。

另外一件事为按钮和检查设置 LayoutParams

       for (int i = 0; i < elements.size(); i++) {

                Button btn = new Button(YourActivity.this); 
                btn.setLayoutParams(new LayoutParams(
                    LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT));
                btn.setGravity(Gravity.CENTER | Gravity.RIGHT);
                btn.setText(elements.get(i)); 
                linearLayout.addView(btn);  
        }

Answer 2

您已通过引用LinearLayout在onCreate()中找到了layout。在方法generateButtonDynamically()

中使用相同的方法

Answer 3

我不知道你是否已经解决了。尝试更改

Button btn = new Button(this);

在

Button btn = new Button(getApplicationContext());

我认为在那一刻这个指向 ExecuteTask 类而不是需要的 Context 。我有类似的问题，我以这种方式解决了它。

Android：在Asyn onPostExecute中动态添加按钮

3 个答案: