Animated backdrop - #657
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This uses jQuery `.animate` to animate the backdrop, which looks really nice. The backdrop is not hidden when moving from one element to another, but is instead moved. Ideally the duration of the animation would be based on whether the backdrop already exists or not (duration 0 if not), but I couldn't find a clean way to check this, somebody else likely will.
houndci-bot
reviewed
Nov 7, 2017
| .height(data.height) | ||
| .width(width - (data.offset.left + data.width)) | ||
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| .animate({top: data.offset.top, left: data.offset.left + data.width, height: data.height, width: width - (data.offset.left + data.width) }, duration) |
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| .offset(top: data.offset.top, left: 0) | ||
| .height(data.height) | ||
| .width(if data.offset.left > 0 then data.offset.left else 0) | ||
| .animate({top: data.offset.top, left: 0, height: data.height, width: Math.max(0, data.offset.left) }, duration) |
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| .offset(top: data.offset.top + data.height, left: 0) | ||
| .height(height - (data.offset.top + data.height)) | ||
| .width(width) | ||
| .animate({top: data.offset.top + data.height, left: 0, height: height - (data.offset.top + data.height), width: width }, duration) |
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| .height(if data.offset.top > 0 then data.offset.top else 0) | ||
| .width(width) | ||
| .offset(top: 0, left: 0) | ||
| .animate({top: 0, left: 0, height: Math.max(0, data.offset.top), width: width }, duration) |
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Author
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p.s. I don't do Coffeescript, so the code is probably pretty bad by CS standards. |
Fix for hide/show issues
houndci-bot
reviewed
Nov 14, 2017
| height: $backdropElement.innerHeight() | ||
| offset: $backdropElement.offset() | ||
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| $('.tour-step-backdrop').not($backdropElement).removeClass('tour-step-backdrop') |
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shaischaudhry
approved these changes
Aug 6, 2022
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This uses jQuery
.animateto animate the backdrop, which looks really nice.The backdrop is not hidden when moving from one element to another, but is instead moved.
Ideally the duration of the animation would be based on whether the backdrop already exists or not (duration 0 if not), but I couldn't find a clean way to check this, somebody else likely will.