since you probably don't want to show the login form after a successful
login,
you could replace the login panel with the logout panel (and vice versa)
this could look something like the following:
// constructor of class LoginPanel
public LoginPanel(final String id) {
super(id);
this.setOutputMarkupId(true);
Form form = new Form("form");
form.add(new AjaxSubmitButton("login", form) {
protected void onSubmit(AjaxRequestTarget target, Form form) {
Panel panel = new LogoutPanel(id);
panel.setOutputMarkupId(true);
LoginPanel.this.replaceWith(panel);
target.addComponent(panel);
}
}.setOutputMarkupId(true));
add(form);
}
any comment on this approach?
On 5/4/07, Francisco Diaz Trepat - gmail <[EMAIL PROTECTED]>
wrote:
Hello, I have two panels, one that is mostly a form for login (user/pass
+button) and the other panel is just a link (onClick==>logou).
Now I changed the button for the login form to an ajax button. Everythig
works fine.
The thing is that I had a code something like this in the logout panel.
@Override
public boolean isVisible(){
return getMyWicketSession().isUserLoggedIn();
}
this code made every round trip evaluate whether the user was still logged
in and if so then render the appropriate option to logout.
how could I add to the target, the logout panel, do I need to pass it
through the constructor :
private final logoutPanel;
public LoginPanel(String id, LogoutPanel panel){
logoutPanel = panel;
}
Ajax....onSubmit(target.......{
target.add(logoutPanel)
}
??
Or could I just set in the constructor of the LogoutPanel
setOutputMarkupId(true) and then call on it from the onSubmit AJAX button?
I am a bit lost on this. Or could I just add a simple JavaScript "
document.getElementById('logoutPanel').style.visibility = "visible";
f(t)
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