I don't believe so. In my app that I'm working on where the .ready isn't
working, I've created a completely new controller and view with nothing in
it except for the example code and the example doesn't work there either on
that.
On Sunday, November 30, 2014 9:28:30 AM UTC-6, Anthony wrote:
>
> Do you accidentally load jQuery twice?
>
> On Sunday, November 30, 2014 9:28:38 AM UTC-5, Jeremiah Peterson wrote:
>>
>> I did load the example in a new dummy app and it works like it should.
>> Any idea what would have changed, or what I could look at in my real app?
>> Could this template I used have messed something up? Thanks for looking.
>>
>> On Sunday, November 30, 2014 6:40:52 AM UTC-6, Niphlod wrote:
>>>
>>> something extremely wrong is going on with your javascript. the "ready"
>>> call is pretty much the default way of handling events once the document
>>> has been loaded.
>>>
>>> On Sunday, November 30, 2014 7:16:04 AM UTC+1, Jeremiah Peterson wrote:
>>>>
>>>> Here's the example from the book, which isn't working for me.
>>>>
>>>> <div class="one" id="a">Hello</div>
>>>> <div class="two" id="b">World</div>
>>>> <script>
>>>> jQuery(document).ready(function(){
>>>> jQuery('.one').click(function(){jQuery('.two').slideToggle()});
>>>> });
>>>> </script>
>>>>
>>>>
>>>> This code does work, just removing the ready.
>>>>
>>>> <div class="one" id="a">Hello</div>
>>>> <div class="two" id="b">World</div>
>>>> <script>
>>>>
>>>> jQuery('.one').click(function(){jQuery('.two').slideToggle()});
>>>>
>>>> </script>
>>>>
>>>> Any ideas?
>>>>
>>>>
>>>>
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