HTML5 Canvas：在画布外拖动

Posted 2023-05-08

【中文标题】HTML5 Canvas：在画布外拖动

【英文标题】：HTML5 Canvas: Dragging outside of canvas

【发布时间】：2012-12-28 13:57:43

【问题描述】：

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 doesn't have this problem，但是我无法识别库中的相关代码。它是如何做到的？

【参考方案1】：

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;
   ...

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，鼠标指针超出该点（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'];
    

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'];
    

在你离开画布时设置 onMouseUp = onMouseOut 坐标并将 isMouseDragged 设置为 false。

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

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/

【讨论】：

和这个想法一样，不幸的是小提琴链接已经死了。

我修改了我的答案（错误的文本格式），并且 jsfiddle 链接现在正在工作......

【参考方案2】：

解决方案是附加到mouseleave 事件，并且不再将用于拖动的按钮考虑在您的逻辑中。