使用路径。它具有适用于少于 21 个 API 的优势(Arc 也因此受到限制,这就是我四边形的原因)。这是一个问题,因为不是每个人都有棒棒糖。但是,您可以指定一个 RectF 并使用它设置值并使用 arc 回到 API 1,但是您将无法使用静态(无需声明新对象来构建对象)。
绘制圆角矩形:
path.moveTo(right, top + ry);
path.rQuadTo(0, -ry, -rx, -ry);
path.rLineTo(-(width - (2 * rx)), 0);
path.rQuadTo(-rx, 0, -rx, ry);
path.rLineTo(0, (height - (2 * ry)));
path.rQuadTo(0, ry, rx, ry);
path.rLineTo((width - (2 * rx)), 0);
path.rQuadTo(rx, 0, rx, -ry);
path.rLineTo(0, -(height - (2 * ry)));
path.close();
作为一个完整的功能:
static public Path RoundedRect(float left, float top, float right, float bottom, float rx, float ry, boolean conformToOriginalPost) {
Path path = new Path();
if (rx < 0) rx = 0;
if (ry < 0) ry = 0;
float width = right - left;
float height = bottom - top;
if (rx > width/2) rx = width/2;
if (ry > height/2) ry = height/2;
float widthMinusCorners = (width - (2 * rx));
float heightMinusCorners = (height - (2 * ry));
path.moveTo(right, top + ry);
path.rQuadTo(0, -ry, -rx, -ry);//top-right corner
path.rLineTo(-widthMinusCorners, 0);
path.rQuadTo(-rx, 0, -rx, ry); //top-left corner
path.rLineTo(0, heightMinusCorners);
if (conformToOriginalPost) {
path.rLineTo(0, ry);
path.rLineTo(width, 0);
path.rLineTo(0, -ry);
}
else {
path.rQuadTo(0, ry, rx, ry);//bottom-left corner
path.rLineTo(widthMinusCorners, 0);
path.rQuadTo(rx, 0, rx, -ry); //bottom-right corner
}
path.rLineTo(0, -heightMinusCorners);
path.close();//Given close, last lineto can be removed.
return path;
}
您希望一直排到那些角位,而不是四边形穿过它们。这就是 conformToOriginalPost 设置为 true 的作用。直接到那里的控制点。
如果你想做这一切但不关心棒棒糖之前的东西,并紧急坚持如果你的 rx 和 ry 足够高,它应该画一个圆圈。
@TargetApi(Build.VERSION_CODES.LOLLIPOP)
static public Path RoundedRect(float left, float top, float right, float bottom, float rx, float ry, boolean conformToOriginalPost) {
Path path = new Path();
if (rx < 0) rx = 0;
if (ry < 0) ry = 0;
float width = right - left;
float height = bottom - top;
if (rx > width/2) rx = width/2;
if (ry > height/2) ry = height/2;
float widthMinusCorners = (width - (2 * rx));
float heightMinusCorners = (height - (2 * ry));
path.moveTo(right, top + ry);
path.arcTo(right - 2*rx, top, right, top + 2*ry, 0, -90, false); //top-right-corner
path.rLineTo(-widthMinusCorners, 0);
path.arcTo(left, top, left + 2*rx, top + 2*ry, 270, -90, false);//top-left corner.
path.rLineTo(0, heightMinusCorners);
if (conformToOriginalPost) {
path.rLineTo(0, ry);
path.rLineTo(width, 0);
path.rLineTo(0, -ry);
}
else {
path.arcTo(left, bottom - 2 * ry, left + 2 * rx, bottom, 180, -90, false); //bottom-left corner
path.rLineTo(widthMinusCorners, 0);
path.arcTo(right - 2 * rx, bottom - 2 * ry, right, bottom, 90, -90, false); //bottom-right corner
}
path.rLineTo(0, -heightMinusCorners);
path.close();//Given close, last lineto can be removed.
return path;
}
所以,
conformToOriginalPost 实际上绘制了一个圆角矩形,底部两位没有圆角。