I have a link which is outside of a Form ... And I want that when user click the link, the Form is submited();
I did : ... onClick(){ form.process(); // But the model of Form Components haven't updated } ... So what I should do for that easy request? -- View this message in context: http://apache-wicket.1842946.n4.nabble.com/New-bie-question-Custom-Form-Submit-tp4392995p4392995.html Sent from the Users forum mailing list archive at Nabble.com. --------------------------------------------------------------------- To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org