如何在asynctask中调用intent?或者如何在onPostExecute中启动新活动?

时间:2016-02-28 09:27:59

标签: android android-intent android-asynctask

有人可以帮助我如何在onPostExecute中调用新活动或在asynctask中添加意图?

我的代码是这样的..

backgroundtask.java

AlertDialog alertDialog;

Context ctx;

BackgroundTask(Context ctx) {
    this.ctx = ctx;
}

@Override
protected void onPreExecute() {
    alertDialog = new AlertDialog.Builder(ctx).create();
    alertDialog.setTitle("Login Information..");
}

@Override
protected String doInBackground(String... params) {
    String reg_url = "http://10.0.2.2/webapp/register.php";
    String login_url = "http://10.0.2.2/webapp/login.php";
    String method = params[0];
    if (method.equals("register")) {

        String name = params[1];
        String user_name = params[2];
        String user_pass = params[3];

        try {
            URL url = new URL(reg_url);
            HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setDoOutput(true);
            OutputStream OS = httpURLConnection.getOutputStream();
            BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(OS, "UTF-8"));
            String data = URLEncoder.encode("name", "UTF-8") + "=" + URLEncoder.encode(name, "UTF-8") + "&" +
                    URLEncoder.encode("user_name", "UTF-8") + "=" + URLEncoder.encode(user_name, "UTF-8") + "&" +
                    URLEncoder.encode("user_pass", "UTF-8") + "=" + URLEncoder.encode(user_pass, "UTF-8");
            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            OS.close();
            InputStream IS = httpURLConnection.getInputStream();

            IS.close();
            return "Registration Success..";

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

    } else if (method.equals("login")) {//EDITED

        String login_name = params[1];
        String login_pass = params[2];

        try {
            URL url = new URL(login_url);
            HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setDoInput(true);
            OutputStream outputStream = httpURLConnection.getOutputStream();
            BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
            String data = URLEncoder.encode("login_name", "UTF-8") + "=" + URLEncoder.encode(login_name, "UTF-8") + "&" +
                    URLEncoder.encode("login_pass", "UTF-8") + "=" + URLEncoder.encode(login_pass, "UTF-8");
            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            outputStream.close();

            InputStream inputStream = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"));
            String response = "";
            String line = "";
            while ((line = bufferedReader.readLine()) != null) {
                response += line;
            }


            bufferedReader.close();
            inputStream.close();
            httpURLConnection.disconnect();
            return response;


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

    }


    return null;
}


@Override
protected void onProgressUpdate(Void... values) {
    super.onProgressUpdate(values);
}

@Override
protected void onPostExecute(String result) {
    if (result.equals("Registration Success..")) {
        Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
        //startActivity(new Intent(this, welcome.class));



    } else {
        alertDialog.setMessage(result);
        alertDialog.show();
    }

}

}

MainActivity.java是我的登录类

EditText ET_NAME, ET_PASS;
String login_name, login_pass;



@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.content_main);

    ET_NAME = (EditText)findViewById(R.id.user_name);
    ET_PASS = (EditText)findViewById(R.id.user_pass);

}
public void userReg(View view) {
    startActivity(new Intent(this,Register.class));

}
public void userLogin(View view) {

    login_name = ET_NAME.getText().toString();
    login_pass = ET_PASS.getText().toString();
    String method = "login";
    BackgroundTask backgroundTask = new BackgroundTask(this);
    backgroundTask.execute(method, login_name, login_pass);
}

我刚接触到android希望你能帮助我..这段代码没有错误btw ..我只是想调用另一个类而不是一个说欢迎的对话框..

5 个答案:

答案 0 :(得分:0)

在你的postExecute方法中。

@Override
protected void onPostExecute(String result) {
    if (result.equals("Registration Success..")) {

        Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
        Intent intent = new Intent(ctx, welcome.class);
        startActivity(intent);

    } else {
        alertDialog.setMessage(result);
        alertDialog.show();
    }
}

答案 1 :(得分:0)

<div class="firstdiv"> <a class="gototwo"> Go to Step 2 </a> </div> <div class="seconddiv" style="display:none"> <a class="gotofirst"> Go back to Step 1 </a> </div>

onPostExecute方法中执行此操作
AsyncTask

它将开启新的活动。

答案 2 :(得分:0)

只需在您的案例中使用 public MainForm() { InitializeComponent(); } private void getInput() { string year; string make; int speed = 0; year = yearTextBox.Text; make = makeTextBox.Text; Car myCar = new Car(year, make); speedLabel.Text = myCar.Speed.ToString(); } private void accelerateButton_Click(object sender, EventArgs e) { getInput(); myCar.Speed(); //error is here myCar does not exist in current context speedLabel.Text = myCar.Speed.ToString(); } private void brakeButton_Click(object sender, EventArgs e) { getInput(); myCar.Brake(); //error is here "myCar does not exist in current context speedLabel.Text = myCar.Brake.ToString(); } } Context

ctx

答案 3 :(得分:0)

使用ctx.startActivity(intent)开始活动。

答案 4 :(得分:0)

您可以使用此

@Override
protected void onPostExecute(String result) {
  if (result.equals("Registration Success..")) {
    Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();

    Intent login = new Intent(ctx, MainActivity.class);
    login.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
    ctx.startActivity(login);
    ((Activity) ctx).finish();


  } else {
    alertDialog.setMessage(result);
    alertDialog.show();
  }

}