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我在使用 HTML5 Canvas 实现拖动功能时遇到问题。我了解缺少场景图等，并设置了一种方法来识别鼠标事件上的拖动矩形。

但是：问题是当鼠标离开画布时会丢失鼠标事件。如果我拖到画布外并释放鼠标按钮，我的代码不会获得 mouseup 事件来相应地改变拖动逻辑。然后，拖动的对象会一直粘在我的鼠标上，直到我再次开始拖动。

fabric.js没有这个问题，但是我无法识别库中的相关代码。它是如何做到的？

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2 回答 2

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首先使用布尔变量 isMouseDragged，它在 mouseDown 之后设置为 true，在 mouseUp 之后设置为 false（或者我稍后会解释的另一个事件）

您可以通过多种方式处理拖动。

if (mouse['isMouseDragged']) {
   mouse['xUp'] = e.clientX - canvas.getBoundingClientRect().left;
   mouse['yUp'] = e.clientY - canvas.getBoundingClientRect().top;
   ...

  1. 在鼠标向上计算您按下和释放按钮的点（将其想象为线）与画布边缘之间的交点，超出该点的是鼠标指针（将其想象为另一条线）。将 mouseUp 位置作为这两条线的交点。

    // FIRST OPTION
if (mouse['mode'] == mouse['INTERSECTION']) {
    if (!mouse['isMouseOver']) {
        var edgeIntersect = mouseDraggedOut(mouse['xDown'], mouse['yDown'], mouse['xUp'], mouse['yUp']);
        mouse['xUp'] = edgeIntersect['x'];
        mouse['yUp'] = edgeIntersect['y'];
    }
}

  2. 将虚拟鼠标指针（坐标）放在画布内。

    // SECOND OPTION
if (!mouse['isMouseOver']) {
    if (mouse['mode'] == mouse['LOCK_INSIDE']) {
        var edgeIntersect = mouseDraggedOut(canvas.width / 2, canvas.height / 2, mouse['x'], mouse['y']);
        mouse['x'] = edgeIntersect['x'];
        mouse['y'] = edgeIntersect['y'];
    }
}

  3. 当您离开画布并将 isMouseDragged 设置为 false 时，使坐标 onMouseUp = onMouseOut。

    // THIRD OPTION
if (mouse['isMouseDragged']) {
   if (mouse['mode'] == mouse['MOUSEOUT_POS']) {
      mouse['xUp'] = mouse['xOut'];
      mouse['yUp'] = mouse['yOut'];
      mouse['isMouseDragged'] = false;
   }
}

要计算画布边缘的相交：

function mouseDraggedOut(x1, y1, x2, y2) {
// x1,y1 = mouseDown;  x2,y2 = mouseUp

var x3, y3, x4, y4;
var thisX, thisY;

if (x2 < 0) {// left edge
    x3 = 0;
    y3 = 0;
    x4 = 0;
    y4 = canvas.height;
    thisX = Math.round(x1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (x2 - x1));
    thisY = Math.round(y1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (y2 - y1));

    // I must do this for other checks, else corners (when two conditions are true) couldn't be handled
    // So I'll handle it one after another
    x2 = thisX;
    y2 = thisY;
}

if (x2 > canvas.width - 1) {// right edge
    x3 = canvas.width - 1;
    y3 = 0;
    x4 = canvas.width - 1;
    y4 = canvas.height - 1;
    thisX = Math.round(x1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (x2 - x1));
    thisY = Math.round(y1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (y2 - y1));
    x2 = thisX;
    y2 = thisY;
}

if (y2 < 0) {// top edge
    x3 = 0;
    y3 = 0;
    x4 = canvas.width - 1;
    y4 = 0;
    thisX = Math.round(x1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (x2 - x1));
    thisY = Math.round(y1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (y2 - y1));
    x2 = thisX;
    y2 = thisY;
}

if (y2 > canvas.height - 1) {// bottom edge
    x3 = 0;
    y3 = canvas.height - 1;
    x4 = canvas.width - 1;
    y4 = canvas.height - 1;
    thisX = Math.round(x1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (x2 - x1));
    thisY = Math.round(y1 + (((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3)) / ((y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1))) * (y2 - y1));
}

return {
    'x' : thisX,
    'y' : thisY
};

}

看源码：http: //jsfiddle.net/WolfeSVK/s2tNr/

于 2013-06-02T19:19:47.183 回答
0

解决方案是附加到mouseleave事件，而不考虑用于拖动的按钮在您的逻辑中不再向下。

于 2018-04-11T11:14:15.913 回答