Question

我想使用asyntask创建httpconnection。将三个参数发布到服务器用户，密码和搜索项。搜索由用户在EditText中提供，这样当用户单击按钮时，搜索项将被发送到服务器。我想在OnclickListener中执行doInbackground（）方法，在listviews上显示服务器的响应。这是AsyncTask类

public class PostToServer extends AsyncTask<String, Void, String> {

@Override
protected void onPostExecute(String result) {

}

@Override
protected String doInBackground(String... arg0) {
     try {
         HttpClient client = new DefaultHttpClient();  
         String postURL = "url";
         String username ="username";
         String password = "password";

         HttpPost post = new HttpPost(postURL);
             List<NameValuePair> params = new ArrayList<NameValuePair>();
             params.add(new BasicNameValuePair("user", username));
             params.add(new BasicNameValuePair("pass", password));
             UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params,HTTP.UTF_8);
             post.setEntity(ent);
             HttpResponse responsePOST = client.execute(post);  
             HttpEntity resEntity = responsePOST.getEntity(); 

             if (resEntity != null) {    
                 Log.i("RESPONSE",EntityUtils.toString(resEntity));
             }
     } catch (Exception e) {
         e.printStackTrace();
     }
    return null;    
}

这是调用click事件的类

public class StartPost extends Activity  {

 @Override
protected void onCreate(Bundle savedInstanceState) {

    super.onCreate(savedInstanceState);
     setContentView(R.layout.result_page);
}
    Button Submit = (Button) findViewById(R.id.btn_search);
     EditText textvalue = (EditText)findViewById(R.id.searcheditText);
     String value = textvalue.getText().toString();
     PostToServer post = new PostToServer();
     CheckInternetConnection check = new CheckInternetConnection(null);

 private OnClickListener click = new OnClickListener() {
     @Override
     public void onClick(final View v) {
         switch(v.getId()){
             case R.id.btn_search:
                 post.execute();
                break;   
         }  
     }
 };
}

问题 1.我做错了什么因为看起来帖子不起作用我怎样才能显示onPostExecute（）的服务器结果？谢谢。

Answer 1

创建一个onPostExecute方法以在textview上显示结果。

protected onPostExecute (String result){
   TextView tv = (TextView) findViewById(R.id.textview1);
   if(result != null){
      tv.setText(result);
   }
}

Android中的AsyncTask HttpConnection

1 个答案: