errrm... thanks Karl...
but I get:
":last" is not a function
obviously, I'm not piecing this together quite right:
var arr = new Array(2)
arr[0] = $('ul#list1').size()
arr[1] = $('ul#list2').size()
arr[2] = $('ul#list3').size()
var largest = arr.sort(function(a,b){return a -
b}).filter(':last')
Karl Swedberg-2 wrote:
>
> Hi Rolfsf,
>
> Instead of .slice(-1), you can use .filter(':last')
>
>
> --Karl
> _________________
> Karl Swedberg
> www.englishrules.com
> www.learningjquery.com
>
>
>
> On Oct 24, 2007, at 3:53 PM, rolfsf wrote:
>
>>
>>
>> Thanks Josh & Andy!
>>
>> I may not be able to upgrade to jQuery 1.2 right away - what's my
>> alternative for slice()?
>>
>> r.
>>
>>
>> Andy Matthews-4 wrote:
>>>
>>>
>>> This will give you the highest val without having to loop over the
>>> array:
>>>
>>> var arr = new Array(6)
>>> arr[0] = "10"
>>> arr[1] = "5"
>>> arr[2] = "40"
>>> arr[3] = "25"
>>> arr[4] = "1000"
>>> arr[5] = "1"
>>> arr[6] = "1001"
>>>
>>> var largest = arr.sort(function(a,b){return a - b}).slice(-1)
>>>
>>>
>>
>> --
>> View this message in context: http://www.nabble.com/help---
>> compare-3-lists%2C-get-the-largest-number-of-items...-
>> tf4685230s27240.html#a13393735
>> Sent from the jQuery General Discussion mailing list archive at
>> Nabble.com.
>>
>
>
>
--
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