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..
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)
在<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();
}
}