Good morning,
I'm a beginner developer on Android.
I created an UI with two EditText and a button.
I want to create an application using twitter (API JTwitter), the the
first stage is to connect to twitter.
On myApp.java I use this:
public class myApp extends Activity implements OnClickListener{
Twitter twitter;
EditText TLogin;
EditText TPassword;
Button Connect;
public void onClick(View src) {
try{
String login = TLogin.getText().toString();
String password = TPassword.getText().toString();
if(!login.equals("") && !password.equals("")){
twitter=new Twitter(login,password);
if(!twitter.isValidLogin()){
TLogin.setText("not valid");
TPassword.setText("not valid");
}
else
twitter.setStatus("Connected");
}
}
catch(Exception e){}
}
public void onCreate(Bundle savedInstanceState) {
try{
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
// find views by id
Connect= (Button) findViewById(R.id.Connect);
TLogin = (EditText) findViewById(R.id.TLogin);
TPassword = (EditText) findViewById(R.id.TPassword);
// Add listener
Connect.setOnClickListener(this);
}
catch(Exception e){}
}
}
When I run this, I get nothing... I can't find where is the problem :s
please help me :s
Else, if I want to call an othor layout using this button, how can I
do it?
Thank you.
--
You received this message because you are subscribed to the Google
Groups "Android Developers" group.
To post to this group, send email to android-developers@googlegroups.com
To unsubscribe from this group, send email to
android-developers+unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/android-developers?hl=en
