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beamish

automatically calculated duration

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Hi,

 

Is it possible when adding a tween to a timeline, to have its' duration be automatically as long as the time gap until the next tween on the timeline (and the last until the total duration of the entire timeline)?  So that if, for example there are 3 tweens at time 0, 7 and 10 each, on a time line with a total duration of 20, the first would have a duration of 7 seconds, the second - 3 seconds, and the last - 10 seconds?  and if the second tween is removed - then the duration of the first tween will automatically update to 10 seconds?...

 

If it's relevant, all tween target the same dom element.

 

Thanks!

 

beamish.

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Possible, yes. Automatic, not really. You would have to create your own function to calculate the duration()s and insertion points.

 

 

The API does not have a method that specifically closes gaps.Methods like getChildren(), , startTime(), duration() and shiftChildren() make it possible to modify a timeline and child tweens virtually any way imaginable. 

 

To remove the 2nd tween and make the first tween long enough to fill up the space between the first tween and beginning of the third tween you could you could write a function that dynamically finds the duration() of the first tween and the startTime() of the third tween and then use the difference to add to the duration() of the first tween.

 

When you add a tween, I'll assume you know where (at what time()) it should be inserted. You could then loop through the children and find the next tween that has a startTime() greater than that insertion time. 

 

All this works very well in theory but it can get complicated fairly quickly depending on how flexible and dynamic this has to be. Hopefully this is enough to get you going in the right direction.

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