(Re-add) spin prop #436
(Re-add) spin prop #436
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Yeah it makes sense for a photo editing type app, I’m good to add the prop will check this out soon |
| let radians = Math.abs((degrees * Math.PI) / 180.0); | ||
| if ((degrees > -45 && degrees <= 45) || (Math.abs(degrees) > 135 && Math.abs(degrees) <= 180)) { | ||
| // Top and Bottom | ||
| x = scaledX * Math.cos(radians); |
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So this is the condition it normally enters and there is no difference between the previous output (x = (e.clientX - rect.left) / zoom;) if spin is 0?
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I will merge and check later anyway just curious
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Since Math.cos(0) = 1, when the spin is 0 that will be evaluated as x = scaledX * 1, so there won't be any difference.
| <div className="ReactCrop__drag-handle ord-s" data-ord="s" /> | ||
| <div className="ReactCrop__drag-handle ord-sw" data-ord="sw" /> | ||
| <div className="ReactCrop__drag-handle ord-w" data-ord="w" /> | ||
| <div className="ReactCrop__drag-handle ord-nw" data-ord="nw" style={this.getRotatedCursor("nw", spin)} /> |
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Does it make sense to change the inline style more than changing the class name?
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My initial thought was to change the inline style, but changing the className might also work.
No, by the looks of it, it should still work the same. Thanks for fixing a redundancy in my code. |
Hi, I apologize for being annoying but I am still having issues using the
rotateprop. Here you can see that when the image rotates, the ReactCrop container stays in place so the image bounds are cropped.I wasn't able to fix it with CSS, so actually I think that I will stick to the behavior that I had before. I am going to call this prop
spinsince it's more like you are "spinning" the whole crop selection around.I know that it might get confusing to distinguish
scale/rotatefromzoom/spin, but basically...zoom/spinare corrections made to the cropping bounds whenever a parent element is transformed.scale/rotateare transformations that are applied directly onto the image.One final note: I had to reset
lastYCrossoverback toyInversedbecause it was giving me problems.