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Avoid setting display: block on dropped cell #130
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- Closes #125, #121, #99 - Fix paradox of choice ($ vs jQuery) - Closes #122 - Update jQueryUI depedency in demos and tests - Closes #130 - Restore display value if set, else use block - Closes #128, #36, #30 - Restore zIndex value always - Closes #117, #110 - Fix destroying widgets prior to initialization - Closes #111 - Allow dropping without ngModel similar to dragging
I think I'm still having the same issue with the latest version 1.0.8. I had a droppable div element with the css property
If I understand this part of the code correctly (I've just really skimmed it so I could easily be wrong), on drop it's replacing the display property with 'block'. If I'm wrong, it's something else in my code. But just wanted to throw this out there in case it's still an issue for anyone else. |
@JesseDahl: Oops, I missed that one. I'll fix it soon. |
- fix(README): Closes #108 and [cleanup](7c06a61) - fix(*): Closes #102 - use $.map over native .map method for IE8 - fix(*): Closes #137 - drag/drop should work even when compile-time debugging is disabled - fix(*): Closes #130 - avoid setting display: block on dropped cell - fix(*): Closes #102 - passing global jQuery reference in case it is renamed with jQuery.noConflict() - fix(*): Closes #143 - introduce deepCopy option to support deep copy on draggable/droppable - perf(*): Undo #55 because some callbacks are performance intensive, especially, onDrag, to run a digest loop for. Call $scope.$apply() within the callback if needed - fix(*): Closes #144 - default or custom confirmation before drop
Hi, i have a table with td elements, which have display: table-cell.
When i drop something on them, the dragndrop module sets them to display: block and they lose the alignment (the table is a calendar).
is it possible to avoid setting it?
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