Sorry but I changed that already cursx is not used anymore.
I advice to use 'onDrag' callback. This way you get element's
coordinates. You check this coordinates against you table position and
each cell position.
let's say: you have a table 500 px width, 500 height, 5 rows and 5
columns, positioned at 100px left and 50px top.Each cell has 100px as
dimension. While you drag the element
you get for instance x=313 and y=214.
To find column int(x-table.left)/cell.width => parseInt(312 - 100)/100
= 3
to find row int(y-table.top)/cell.height => parseInt(214 - 50)/100 = 1
remember this last cell from row 1 and column 3 and when the dragging
stops you know that the element was dropped on that last cell.
Mark Gibson wrote:
> Mark Gibson wrote:
>
>> Ideally I'd like to have a single Droppable on the whole table,
>> and use the 'ondrop' callback, but having the position of the pointer
>> (relative to the table element) passed to the callback function,
>> is this possible?
>>
>
> Right, I've solved it. To recap the problem:
>
> I have a palette of items, which need associating with columns in a
> table. I wanted to drag the item into the table (using ghosting drag).
> On dropping the item it would be removed from the palette and placed
> in the TH cell of whichever column it was dropped in.
>
> First I tried adding a Droppable to every TH and TD - this caused a
> major delay at the beginning of the drag operation.
>
> Now, I've got a Droppable on just the TABLE:
>
> $("table").Droppable({
> accept: 'item',
> tolerance: 'pointer',
> ondrop: function(item) {
> var dropx = item.dragCfg.cursx; // The cursor position at time
> of drop
>
> var ths = $(this).find('th'); // Get all heading cells
>
> // Find the column in which the drop occurred
> for (var i=1; i < ths.length; i++) {
> if (dropx < jQuery.iUtil.getPosition(ths[i]).x) break;
> }
>
> // Place the item in the column heading
> $(item).remove().appendTo(ths[i-1]);
> }
> });
>
> I have to ask though, if using of item.dragCfg.cursx is future proof?
> Is there a better/safer way I should be doing this?
>
> Otherwise, I hope this may help someone else.
>
> Cheers
> - Mark
>
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