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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