HTML5：从画布弧中获取点击位置

Question

查看此jsFiddle 帖子以获取工作弧图；感谢Simon Sarris 修复了我之前的问题。

我正在使用KineticJS 插件来创建形状并使用事件处理程序。假设您单击了弧上的某个位置并且弧知道您单击的位置（x、y），那么这两个坐标如何用于确定百分比？

当您点击任意位置时，总百分比始终为 100%。

插件

为了更简单，我可以对 (x, y) 做些什么来虚拟弯曲对象，使 x 从 0 变为最大 x？

Answer 1

简单的三角函数。 sin(angle) = opposite / adjacent。 opposite 是 y 值，adjacent 是 x 值。所以Math.asin((xx - x) / (yy - y)) 其中 xx 和 yy 是圆弧中心的坐标。这为您提供了角度，然后您可以将其除以 2 * Math.PI。

我不记得负数会发生什么。您可能需要获取参数的Math.abs 值，然后计算出点击在哪个象限（使用< 和> 很容易做到）并为每个象限添加Math.PI / 2。

Answer 2

这包括检查鼠标是否在弧内：

// Return range is 0 to Math.PI * 2
function get_mouse_circle_angle(origin_x, origin_y, mouse_x, mouse_y) {
    var mouse_angle = Math.atan2(mouse_y - origin_y, mouse_x - origin_x);
    if (mouse_angle < 0) {
        mouse_angle = (Math.PI * 2) + mouse_angle;
    }
    return mouse_angle;
}

// Return range is [0, 1)
// 0/1 is 3 oclock
function get_mouse_circle_percent(origin_x, origin_y, mouse_x, mouse_y) {
    var mouse_angle = get_mouse_circle_angle(origin_x, origin_y, mouse_x, mouse_y);
    return mouse_angle / (2 * Math.PI);
}

function get_mouse_arc_pos(origin_x, origin_y, mouse_x, mouse_y, radius, thickness) {
    var mouse_angle = Math.atan2(mouse_y - origin_y, mouse_x - origin_x);
    if (mouse_angle < 0) {
        mouse_angle = (Math.PI * 2) + mouse_angle;
    }
    var mouse_percent = mouse_angle / (2 * Math.PI);

    var circle_edge_x = origin_x + (radius + thickness / 2) * Math.cos(mouse_angle);
    var circle_edge_y = origin_y + (radius + thickness / 2) * Math.sin(mouse_angle);

    var arc_inside_x = origin_x + (radius - thickness / 2) * Math.cos(mouse_angle);
    var arc_inside_y = origin_y + (radius - thickness / 2) * Math.sin(mouse_angle);

    var is_in_circle = true;

    if (mouse_angle <= (2 * Math.PI) * 0.25) {
        if (mouse_x > circle_edge_x || mouse_y > circle_edge_y)
            is_in_circle = false;
    }
    else if (mouse_angle <= (2 * Math.PI) * 0.5) {
        if (mouse_x < circle_edge_x || mouse_y > circle_edge_y)
            is_in_circle = false;
    }
    else if (mouse_angle <= (2 * Math.PI) * 0.75) {
        if (mouse_x < circle_edge_x || mouse_y < circle_edge_y)
            is_in_circle = false;
    }
    else {
        if (mouse_x > circle_edge_x || mouse_y < circle_edge_y)
            is_in_circle = false;
    }

    var is_in_arc = is_in_circle;
    if (is_in_circle) {
        if (mouse_angle <= (2 * Math.PI) * 0.25) {
            if (mouse_x < arc_inside_x || mouse_y < arc_inside_y)
                is_in_arc = false;
        }
        else if (mouse_angle <= (2 * Math.PI) * 0.5) {
            if (mouse_x > arc_inside_x || mouse_y < arc_inside_y)
                is_in_arc = false;
        }
        else if (mouse_angle <= (2 * Math.PI) * 0.75) {
            if (mouse_x > arc_inside_x || mouse_y > arc_inside_y)
                is_in_arc = false;
        }
        else {
            if (mouse_x < arc_inside_x || mouse_y > arc_inside_y)
                is_in_arc = false;
        }
    }

    return {
        angle: mouse_angle,
        percent: mouse_percent,
        is_in_circle: is_in_circle,
        is_in_arc: is_in_arc
    };
}

Answer 3

没有真正测试过，但从技术上讲，它应该可以工作：

// Where x1 and y1 should be the coordinates of the arc's center
function angle(x1, y1, x2, y2) {
    // Calculate a · b
    var nominator = x1 * x2 + y1 * y2;

    // Calculate ||a|| ||b||
    var denominator = Math.sqrt(x1*x1 + y1*y1) * Math.sqrt(x2*x2 + y2*y2);
    if (denominator == 0) return 0; // Indifinite angle

    // Return the angle
    return Math.acos(nominator / denominator);
}

// Returns a percent, might be negative
var percent = angle(0, 0, mouseX, mouseY) / (2*Math.PI);

编辑

对于负数，您可以尝试加 1，因为它在 [-1, 1] 范围内

if (percent < 0) percent += 1;