Question

paper.rect(0, 0, settings.width, settings.height, settings.radius);

创建一个带圆角的漂亮矩形。是否可以创建一个只有一个圆角的矩形？

Answer 1

如果您使用Raphael JS：

Raphael.fn.roundedRectangle = function (x, y, w, h, r1, r2, r3, r4){
  var array = [];
  array = array.concat(["M",x,r1+y, "Q",x,y, x+r1,y]); //A
  array = array.concat(["L",x+w-r2,y, "Q",x+w,y, x+w,y+r2]); //B
  array = array.concat(["L",x+w,y+h-r3, "Q",x+w,y+h, x+w-r3,y+h]); //C
  array = array.concat(["L",x+r4,y+h, "Q",x,y+h, x,y+h-r4, "Z"]); //D

  return this.path(array);
};

要一个只有右上角圆角的矩形

var paper = Raphael("canvas", 840, 480);
paper.roundedRectangle(10, 10, 80, 80, 0, 20, 0, 0);

来源和在线示例：http://www.remy-mellet.com/blog/179-draw-rectangle-with-123-or-4-rounded-corner/

Answer 2

很老的问题，这是一条更好的道路。我将它转换为相对坐标，这在动画中应该更好......

Raphael.fn.roundedRectangle = function (x, y, w, h, r1, r2, r3, r4){
  var array = [];
  array = array.concat(["M",x+r1,y]);
  array = array.concat(['l',w-r1-r2,0]);//T
  array = array.concat(["q",r2,0, r2,r2]); //TR
  array = array.concat(['l',0,h-r3-r2]);//R
  array = array.concat(["q",0,r3, -r3,r3]); //BR
  array = array.concat(['l',-w+r4+r3,0]);//B
  array = array.concat(["q",-r4,0, -r4,-r4]); //BL
  array = array.concat(['l',0,-h+r4+r1]);//L
  array = array.concat(["q",0,-r1, r1,-r1]); //TL
  array = array.concat(["z"]); //end

  return this.path(array);
};

Raphael-JS：有一个圆角的Rect

3 个答案: