如何在自定义alertdialog中启动活动

Question

我在Activatedialog注册＆gt;活动中使用此onPostExecute方法，并且当警告对话框消失时需要启动login活动..但是我试图这样做，但我得到错误。

 public void Activatedialog(String jsonmsg, final String name)
{

    final AlertDialog.Builder alert = new AlertDialog.Builder(ct);

    alert.setTitle("Activate Account");
    alert.setMessage(jsonmsg);

    // Set an EditText view to get user input 

    final EditText input = new EditText(ct);
    alert.setView(input);
    alert.setCancelable(true);
    Log.d("MYLOG","before +ve button click");

    alert.setPositiveButton("Ok", new DialogInterface.OnClickListener()
    {
            public void onClick(DialogInterface dialog, int whichButton) 
            {
                      String code = input.getText().toString();

                      Log.d("MYLOG","before try");
                      try{
                            List<NameValuePair> params = new ArrayList<NameValuePair>();
                            params.add(new BasicNameValuePair("u",name ));
                            params.add(new BasicNameValuePair("code",code ));
                            JSONObject json = jsonParser.makeHttpRequest(ACTIVATION_URL,"POST", params);

                            Log.d("MYLOG","after request for json");

                            stcode = json.getJSONObject("activate").getInt("statuscode");
                            stmsg = json.getJSONObject("activate").getString("statusmessage");

                            Log.d("MYLOG","after json parsing : json data is "+stmsg+"stcodeis "+stcode);

                            if(stcode == 1)
                            {
                                dialog.dismiss();
                                Log.d("MYLOG"," in check json data is "+stmsg+" stcodeis "+stcode);

                                Intent i = new ("LOGIN");startActivity(i);![i get this error][1]

                                Toast.makeText(ct, " "+stmsg, Toast.LENGTH_SHORT).show();
                            }
                            else if(stcode == 0)
                            {
                                Log.d("MYLOG"," in check json data is "+stmsg+" stcodeis "+stcode);
                                Toast.makeText(ct, " "+stmsg+" "+stcode, Toast.LENGTH_SHORT).show();

                            }
                      }
                      catch(JSONException e)
                      {
                          e.printStackTrace();
                          Toast.makeText(ct, ""+e, Toast.LENGTH_LONG).show();
                      }
            }
    });
    alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener()
    {
              public void onClick(DialogInterface dialog, int whichButton)
              {
                        // Canceled.
                        Toast.makeText(ct, "Your Account is not Activated.Login to Activate", Toast.LENGTH_LONG).show();
              }
    });

    alert.show();

}

 }

Answer 1

要启动活动，意图必须具有以下参数。上下文和目标活动。让我们开始活动“Login.java”。

Intent i = new Intent(getApplicationContext(), Login.class);
startActivity(i);

Answer 2

您可以这样做：

AlertDialogManager alert = new AlertDialogManager(StartActivity.this);

AlertDialogManager构造函数：

public void AlertDialogManager(final Context context) {

    activity = (Activity) context;

}

你可以启动这样的活动：

Intent intent = new Intent(activity , Login.class);
    activity.startActivity(intent);

Answer 3

您错误地创建了意图。

1＆GT;通过向项目添加新活动，添加您要打开的相关活动Login。

2＆gt;按以下方式创建意图

Intent intent = new Intent(getApplicationContext(), Login.class);

3＆gt;使用创建的意图启动活动

startActivity(i);