Your function is fine, however when you *call* the function removeItem,
you'll need to pass in two arguments -- otherwise obj will be undefined:
removeItem('myitem'); // obj is undefined
removeItem('myitem', myobject); // obj is defined
-- Josh
----- Original Message -----
From: "Mark" <[EMAIL PROTECTED]>
To: <jquery-en@googlegroups.com>
Sent: Thursday, June 19, 2008 4:02 PM
Subject: [jQuery] Why do i keep getting: test is not defined!
hey,
function removeItem(name, obj)
{
var answer = confirm("Are you sure you want to delete: " + name + "?")
var test = obj;
if (answer)
{
$(test.parentNode.parentNode).fadeOut("slow");
}
}
the code is simple and still not working.. the issue is that i try to
call a variable in a if that wasn't made there.. but how can i resolve
that?
Why is javascript not working how you would expect it to work :S i
know Java and PHP well and things like this are no issue in them.
o and in this code i already put obj in test.. i've tried it with obj
first but that gave the same error as in the title.
Thanx.
