Reputation: 65
I've got this totally simple code for changing one image (#form1)
with another (#form2)
when clicking on #third
, which doesn't seem to work.
I've thought maybe it's because I've already used $(document).ready
before but now I'm reading it's fine to use it several times.
So what is wrong? Could someone help me out?
My code:
$(document).ready(function(){
$("#form2").hide();
$("#third").click(function() {
$("#form1").replaceWith("#form2");
});
});
Upvotes: 1
Views: 1448
Reputation: 16764
Because you replace form with a string not with an object
$("#form1").replaceWith($("#form2"));
Upvotes: 3
Reputation: 64526
Pass the element to .replaceWith()
, not the selector string:
$("#form1").replaceWith($("#form2"));
// ^^ ^ was missing
The argument to .replaceWith()
can be a HTML string, DOM element or jQuery element, but it should never be the selector string like you have.
Upvotes: 3