How about overriding $.end() in the grid plugin, so that it returns the
original jquery object if it's run on the grid controller?
Of course, you'd have to remember to make $.end() behave normally if run
elsewhere. Or, you could just create an $.endGrid() function for the
controller, and not worry about messing with $end() at all.
- Brian
> Hi,
>
>> Controller set of methods is returned..
>>
>> $('#grid').grid().data(data).drig().show()
>> $('#grid').grid().scrollToRow(6).drig().css("border", "1px")
>>
>> A controller object is returned..
>>
>> var grid = null
>> $("grid").grid({
>> data: data,
>> onComplete: function(controller) { grid = controller }
>> })
>> grid.srollToRow(6)
>>
>> What is best practice?
>
> How about the Idea to have a jQuery-function that returns the controller
> object:
>
> $('#grid').grid({data:data}).show().gridController().scrollToRow(6);
>
> That way the function that creates the grid does not break the chain, but
> there is a function that returns a controller object. You don't
> necessarily
> need a way to come back to jQuery if you do it that way, because you can
> chose to get the controller Object later in the queue.
>
> You should think if you want a controller object that can handle mutliple
> grids at the same time:
>
> $('.allmygrids').grid({data:data}).show().gridController().scrollToRow(6);
>
> If not, then gridController() needs to take a parameter to know which of
> the
> grids you mean:
>
> $('.allmygrids').grid({data:data}).show().gridController(42).scrollToRow(6);
>
> Christof
>
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