Question

我在主活动上有一个登录按钮，执行aysncTask，启动另一个活动onPostExecute，但是当用户多次点击按钮时它多次启动活动，我该如何防止这种情况发生？

这是我的onClick Listener：

loginButton.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {

                if (IP != "null") {
                    password = enterPassword.getText().toString();
                    progressB.setVisibility(View.VISIBLE);
                new httpLogin().execute();}
                else {
                    Toast.makeText(getApplicationContext(), "please enter H2O host address before attempting to login", Toast.LENGTH_LONG).show();
                }
            }
        });

这是AsyncTask代码：

public class httpLogin extends AsyncTask<Void,Void,String> {
        String rs,result;
        JSONArray jArray;
        boolean r,isConnected;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            progressB.setProgress(0);
            isConnected=isOnline();
        }

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

            try {

                String link = "http://"+IP+"/H2O.asmx/Check_Waiter?Password="+password;
                HttpClient client = new DefaultHttpClient();
                HttpGet request = new HttpGet();
                request.setURI(new URI(link));
                HttpResponse response = client.execute(request);
                BufferedReader reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"), 8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                result = sb.toString();
                String[] separated = result.split(">");
                JSONObject jObject = new JSONObject(separated[2]);
                jArray = jObject.getJSONArray("Waiter");
                if(jArray.length()>0) {
                    JSONObject json_data = jArray.getJSONObject(0);
                    waiterID = json_data.getString("UserNo");
                    waiterName = json_data.getString("UserName");
                    r=true;
                }
                else r=false;
                rs = "sucessful";
            } catch (Exception e) {

                rs = "Fail";
            }
            return rs;
        }
        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            if(isConnected)
            if (r) {
                Toast.makeText(getApplicationContext(), "User ID:" + waiterID + "\n Username:" + waiterName, Toast.LENGTH_SHORT).show();

               Intent intentFSelector = new Intent(MainActivity.this, FloorSelector.class);

        startActivity(intentFSelector); 
            } 
    }

这是ping方法，用于检查本地服务器是否可用：

public boolean isOnline() {
        String[] separated = IP.split(":");
        String hostip=separated[0];
        Runtime runtime = Runtime.getRuntime();
        try {

            Process ipProcess = runtime.exec("/system/bin/ping -c 1 "+hostip);
            int     exitValue = ipProcess.waitFor();
            return (exitValue == 0);

        } catch (IOException e)          { e.printStackTrace(); }
        catch (InterruptedException e) { e.printStackTrace(); }

        return false;
    }

Answer 1

执行AsyncTask后禁用该按钮。

if (IP != "null") {
    password = enterPassword.getText().toString();
    progressB.setVisibility(View.VISIBLE);
    new httpLogin().execute();
    ((Button) view).setEnabled(false);
}

理想情况下，您应该立即拨打startActivity()按钮点击＆amp;执行AsyncTask中的onCreate()。在下载数据时，您将在其中一个数据视图中显示“loading ...”。完成后，您将更新onPostExecute()中的视图，或者如果下载失败则显示错误消息;说用户是否离线。

Answer 2

试试这个并插入你的意图：

intent.addFlags(Intent.FLAG_ACTIVITY_REORDER_TO_FRONT);

Answer 3

您可以使用ProgressDialog避免多次点击登录按钮。

ProgressDialog authProgressDialog;

在AsyncTask中

public class httpLogin extends AsyncTask<Void,Void,String> {
    String rs,result;
    JSONArray jArray;
    boolean r,isConnected;

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        authProgressDialog = ProgressDialog.show(MainActivity.this, "", "Loading....", true, false);   
    }

protected Object doInBackground(Object... params) {
      //your code
    }

 protected void onPostExecute(Object result){

            if (authProgressDialog!=null) {
                if (authProgressDialog.isShowing()) {
                        authProgressDialog.dismiss();       
                }
            }
    }
   }

单击按钮时将显示进度对话框，执行AsyncTask后该对话框将关闭

多次单击按钮时反复启动的活动

3 个答案: