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在我的上一篇文章中，有人告诉我我的结构一团糟，所以我已经改变了它，所以我不想从一个活动中做所有事情。

我的 loginactivity 获取登录详细信息，然后调用 asycn 任务来检查用户是否存在于我的外部数据库中：

public class LogIn extends Activity {


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        setContentView(R.layout.activity_login);


    }

    public void logIn(View view){
        EditText textU;
        EditText textP;

        //get userName and password from edit text
        textU   = (EditText)findViewById(R.id.Username);
        textP   = (EditText)findViewById(R.id.Password);

        String userName = textU.getText().toString();
        String passW = textP.getText().toString();

        Log.d("logIN", userName);
        Log.d("logIN", passW);


        //log in url
        String url = "myURL";
        String userURLComp = "u=" + userName;
        String pURLComp = "&p=" + passW;

        url = url + userURLComp + pURLComp;

        Log.d("URL", url);

        //async task for getting json
        new ReadLogInJSON(this).execute(url);





    }




}

我的问题是，在异步任务检查用户是否在数据库中之后，我不知道如何从异步任务启动新活动：

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

    Context c;

    public ReadLogInJSON(Context context)
    {
         c = context;
    }

    @Override
    protected String doInBackground(String... arg0) {
        // TODO Auto-generated method stub
        return readJSONFeed(arg0[0]);
    }

    protected void onPostExecute(String result){

        //decode json here
        try{

            JSONObject json = new JSONObject(result);
            String status = json.getString("status");

            if(status.equals("no")){
                //toast logIN failed

                String message = "Log In Failed";
                Toast.makeText(c,  message, Toast.LENGTH_SHORT).show();

            }

            else{
                //get userName
                String userName = json.getString("userName");

                //get user ID
                String userID = json.getString("userID");


                //set preferences
                SharedPreferences preferences = PreferenceManager.getDefaultSharedPreferences(c);
                SharedPreferences.Editor editor = preferences.edit();
                editor.putString("userName",userName);
                editor.putString("userID",userID);
                editor.commit();

                //launch new activity



            }

        }
        catch(Exception e){

        }

    }

    public String readJSONFeed(String URL) {
        StringBuilder stringBuilder = new StringBuilder();
        HttpClient httpClient = new DefaultHttpClient();
        HttpGet httpGet = new HttpGet(URL);
        try {
            HttpResponse response = httpClient.execute(httpGet);
            StatusLine statusLine = response.getStatusLine();
            int statusCode = statusLine.getStatusCode();
            if (statusCode == 200) {
                HttpEntity entity = response.getEntity();
                InputStream inputStream = entity.getContent();
                BufferedReader reader = new BufferedReader(
                        new InputStreamReader(inputStream));
                String line;
                while ((line = reader.readLine()) != null) {
                    stringBuilder.append(line);
                }
                inputStream.close();
            } else {
                Log.d("JSON", "Failed to download file");
            }
        } catch (Exception e) {
            Log.d("readJSONFeed", e.getLocalizedMessage());
        }        
        return stringBuilder.toString();
    }

}

更新：

我试着用这个启动它：

c.startActivity(new Intent(c, Search.class));

但没有启动。

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1 回答 1

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首先，您确实应该在逻辑上为您的类名添加后缀，例如“LoginAsyncTask”和“SearchActivity”。其次，是的，启动活动的正确（也是唯一）方法是调用startActivity(Intent):) 尝试在之前和之后添加日志startActivity，以验证是否onPostExecute真的运行。

于 2013-06-22T14:37:57.263 回答