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\foreach 带有曲线的索引名称

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\foreach 带有曲线的索引名称

我正在构建几条曲线，它们的构造方式各不相同，但它们的路径名称为=lambda1，...，lambda4。后来，我将它们与一条路径相交。附件是一个可行最小示例。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
\path[name path=t1] (6,0) -- (6,7); 
\draw[name path=lambda1] (0,4) -- (7,3.5); 
\draw[name path=lambda2](0,2.5) sin (7,4.5); 
\draw[name path=lambda3] (3,0) parabola (7,5); 
\draw[name path=lambda4] (2,0) .. controls (3,4) .. (7,7);
\fill [red, name intersections={of=lambda1 and t1}] (intersection-1) circle (2pt) node    [below] {$\lambda_1$};
\fill [red, name intersections={of=lambda2 and t1}] (intersection-1) circle (2pt) node[below] {$\lambda_2$}; 
\fill [red, name intersections={of=lambda3 and t1}] (intersection-1) circle (2pt) node[below] {$\lambda_3$}; 
\fill [red, name intersections={of=lambda4 and t1}] (intersection-1) circle (2pt) node[below] {$\lambda_4$};
\end{tikzpicture}
\end{document}

最后 4 行需要一个 \foreach，但我无法让它完全工作。对我来说，似乎

\foreach \n in {1,...,4} {
\fill [red, name intersections={of=lambda\n and t1}] 
(intersection-1) circle (2pt) node[below] {$\lambda_\n$};}

应该是正确的解决方案，但事实并非如此。然而，

\foreach \n in {1,...,4} {
\fill[red, name intersections={of=t1 and lambda\n}] 
(intersection-1) circle (2pt) node[below] {$\lambda_\n$};}

运行良好。我这里遗漏了什么？

答案1

如果lambda\n通过将其括在内进行分组{...}，则它可以工作：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}
\path[name path=t1] (6,0) -- (6,7); 
\draw[name path=lambda1] (0,4) -- (7,3.5); 
\draw[name path=lambda2](0,2.5) sin (7,4.5); 
\draw[name path=lambda3] (3,0) parabola (7,5); 
\draw[name path=lambda4] (2,0) .. controls (3,4) .. (7,7);
\foreach \n in {1,...,4} {
    \fill [blue, name intersections={of={lambda\n} and t1}] 
        (intersection-1) circle (2pt) node[below] {$\lambda_\n$};
} 
\end{tikzpicture}
\end{document}

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