Tikz 圆圈边缘颜色不显示

Question

我正在尝试绘制一些带有黑色边缘和白色表面颜色的圆圈。这些圆圈位于椭圆的边缘。我不明白为什么此代码不起作用：

\documentclass{minimal}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
    \shadedraw [inner color=black!10,outer color=black!60,domain=0:2*pi] plot ({2*cos(\x r)},{1*sin(\x r)});
    \foreach \a in {0,0.6283,...,6.2832}
           {
               % I was expecting circles with a black edge
               \fill [black,fill=white] ({2*cos(\a r)},{1*sin(\a r)}) circle (4pt);
           }
\end{tikzpicture}
\end{document}

这是我得到的（圆圈边缘要么是白色的，要么根本没有画出来）：

谢谢！

Answer 1

使用draw=black或\filldraw。正如 Grimler 所说，\fill颜色仅用作填充颜色。您没有告诉 TikZ 绘制形状，只是填充它，并且填充颜色设置为白色。通过添加，draw=black形状也会使用指定的颜色绘制。\filldraw[black,fill=white]形状的轮廓用黑色绘制，并用白色填充。（实际上，顺序是先填充，然后绘制。）

\documentclass{article}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
    \shadedraw [inner color=black!10,outer color=black!60,domain=0:2*pi] plot ({2*cos(\x r)},{1*sin(\x r)});
    \foreach \a in {0,0.6283,...,6.2832}
           {
           % \filldraw [black,fill=white] ({2*cos(\a r)},{1*sin(\a r)}) circle (4pt);
               \fill [draw=black,fill=white] ({2*cos(\a r)},{1*sin(\a r)}) circle (4pt);
           }
\end{tikzpicture}
\end{document}

Answer 2

或者使用\filldraw。也许您想剪辑更多，samples以便获得更平滑的椭圆。

\documentclass{article}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}[,line width=1.4\pgflinewidth]
    \clip [domain=0:2*pi,samples=500] plot ({2*cos(\x r)},{1*sin(\x r)});
    \shadedraw [inner color=black!10,outer color=black!60,domain=0:2*pi,samples=500] plot ({2*cos(\x r)},{1*sin(\x r)});
    \foreach \a in {0,0.6283,...,6.2832}
           {
               % I was expecting circles with a black edge
               \filldraw [black,fill=white,line width=0.7\pgflinewidth] ({2*cos(\a r)},{1*sin(\a r)}) circle (4pt);
           }
\end{tikzpicture}
\end{document}