I have a password input component (id=password) and a login button (id = bt_login). what I want is when users press "Enter Key", the action for login button is activated.
I tried this, if (event.keyCode == 13) {this.form1.bt_login.click();}
, but it does not work.
What should I do?
Thanks in advance,
I tried something similar to http://blogs.sun.com/roller/page/divas?entry=turning_an_input_component_into
I created a page with a Password Field component and a Button component.
Using Page Navigation, I created a link from the Button to Page2 and named it "go".
Back in the Visual Designer, I right-clicked in the Password Field and chose Auto-Submit on Change.
I double-clicked the button (which had an id of goButton) and changed its code as follows:
public String goButton_action() {
// TODO: Process the button click action. Return value is a navigation
// case name where null will return to the same page.
log("go button");
return "go";
}
That way I could tell if the action method was executed.
Then I went back to the Visual Designer and double-clicked the Password Field component. The IDE created a process value change method and displayed the method in the source editor. I changed it as follows:
public void passwordField1_processValueChange(ValueChangeEvent event) {
Application application = getApplication();
NavigationHandler navigator = application.getNavigationHandler();
FacesContext facesContext = getFacesContext();
String result = goButton_action();
try {
navigator.handleNavigation(facesContext, null, result);
} catch (Exception e) {
log("can't navigate to " + (String) result);
}
}
I clicked the Servers window header to display the window, right-clicked Deployment Server and chose View Server Log.
I ran the app. I typed something in the Password Field and pressed Enter. I looked at the log and saw "go button".
Hope this helps,
Chris
