svg/d3.js在矩形的一角上的圆角

Question

我知道svg有一个内置功能来做圆角,但我需要在四个角中的两个角上做圆角.

我知道我可以在彼此之上绘制多个矩形来模仿它,但这看起来有点俗气.有什么方法可以使用剪辑或任何d3.js方法吗？

现在我有一个水平条形图,其形状如下:

    rects.enter().append("rect")
        .attr("x",function(d,i) { return x(0); })
        .attr("width",function(d) { return x(d.value) - x(0); })
        .attr("height",y.rangeBand())
        .attr("y",function(d) { return y(d.name); })

我试图在rect的右侧产生圆角,但不知道该怎么做.

Answer 1

扩展@ robert-longson的答案,您可以使用SVG的椭圆弧命令来制作角点,并结合直线边缘的lineto命令.这些与路径元素一起使用.这是一个可能的实现:

// Returns path data for a rectangle with rounded right corners.
// The top-left corner is ?x,y?.
function rightRoundedRect(x, y, width, height, radius) {
  return "M" + x + "," + y
       + "h" + (width - radius)
       + "a" + radius + "," + radius + " 0 0 1 " + radius + "," + radius
       + "v" + (height - 2 * radius)
       + "a" + radius + "," + radius + " 0 0 1 " + -radius + "," + radius
       + "h" + (radius - width)
       + "z";
}

然后,您可以调用此函数来计算"d"属性.例如:

rects.enter().append("path")
    .attr("d", function(d) {
      return rightRoundedRect(x(0), y(d.name), x(d.value) - x(0), y.rangeBand(), 10);
    });

实例:

• http://bl.ocks.org/3468167

可选:如果您愿意,可以重构righ​​tRoundedRect函数以使其可配置,而不是使用大量参数.这种方法类似于D3的内置形状发生器.例如,您可以使用如下的rect生成器:

rects.enter().append("path")
    .attr("d", rightRoundedRect()
      .x(x(0))
      .y(function(d) { return y(d.name); })
      .width(function(d) { return x(d.value) - x(0); })
      .height(y.rangeBand())
      .radius(10));

有关该方法的更多详细信息,请参阅可配置功能教程.

Answer 2

只是为了扩展给出的答案,这里有一个更全面的函数来返回你的rect的路径.

x: x-coordinate
y: y-coordinate
w: width
h: height
r: corner radius
tl: top_left rounded?
tr: top_right rounded?
bl: bottom_left rounded?
br: bottom_right rounded?

function rounded_rect(x, y, w, h, r, tl, tr, bl, br) {
    var retval;
    retval  = "M" + (x + r) + "," + y;
    retval += "h" + (w - 2*r);
    if (tr) { retval += "a" + r + "," + r + " 0 0 1 " + r + "," + r; }
    else { retval += "h" + r; retval += "v" + r; }
    retval += "v" + (h - 2*r);
    if (br) { retval += "a" + r + "," + r + " 0 0 1 " + -r + "," + r; }
    else { retval += "v" + r; retval += "h" + -r; }
    retval += "h" + (2*r - w);
    if (bl) { retval += "a" + r + "," + r + " 0 0 1 " + -r + "," + -r; }
    else { retval += "h" + -r; retval += "v" + -r; }
    retval += "v" + (2*r - h);
    if (tl) { retval += "a" + r + "," + r + " 0 0 1 " + r + "," + -r; }
    else { retval += "v" + -r; retval += "h" + r; }
    retval += "z";
    return retval;
}

Answer 3

如果其他人最终想要对元素的所有角进行舍入rect,则可以rx向rect元素添加属性(如上面提到的@mbostock中提到的):

var rectangle = group.append("rect")
    .attr("width", 60)
    .attr("height", 75)
    .attr("rx", 4)
    .style("fill", function(d) { return "#e6653e"; })
    .style("stroke", function(d) { return d3.rgb("#e6653e").darker(); })

Answer 4

任何寻找 Eslinted 版本的 stackmate -s 的人都会得到答案：

function roundedRect(x, y, w, h, r, tl, tr, bl, br) {
  let retval;
  retval = `M${x + r},${y}`;
  retval += `h${w - (2 * r)}`;
  if (tr) {
    retval += `a${r},${r} 0 0 1 ${r},${r}`;
  } else {
    retval += `h${r}`; retval += `v${r}`;
  }
  retval += `v${h - (2 * r)}`;
  if (br) {
    retval += `a${r},${r} 0 0 1 ${-r},${r}`;
  } else {
    retval += `v${r}`; retval += `h${-r}`;
  }
  retval += `h${(2 * r) - w}`;
  if (bl) {
    retval += `a${r},${r} 0 0 1 ${-r},${-r}`;
  } else {
    retval += `h${-r}`; retval += `v${-r}`;
  }
  retval += `v${((2 * r) - h)}`;
  if (tl) {
    retval += `a${r},${r} 0 0 1 ${r},${-r}`;
  } else {
    retval += `v${-r}`; retval += `h${r}`;
  }
  retval += 'z';
  return retval;
}