On Thu, Nov 6, 2014 at 7:50 PM, Fabrice <couvreur.fabr...@wanadoo.fr> wrote:
> Hi Hans,
> It's true, I'm a little confused. In fact, the problem is the \
> starttikzpicture \ endtikzpicture.
>
\usemodule[tikz]
\usetikzlibrary[calc]
\usetikzlibrary[3d]
\usetikzlibrary[arrows.meta]
\usetikzlibrary[positioning]
\definecolor[H1prime][c=0.50, m=0.00, y=1.00, k=0.00]
\definecolor[H4][c=0.07, m=0.00, y=0.15, k=0.00]
\tikzset{%
cube join/.style={
thick, -{Stealth},
},
cube face/.style={
minimum size=1cm, outer sep=0pt,
draw=white, thick, line join=round,
shading=ball,
text=white,
},
face color/.style={cube face/.append style={ball color=#1}},
pics/cube/.style args={#1 with #2}{
code={
\node [cube face, label={[name=-label]below:#2}] (-front) {#1};
\node [cube face] (-top) at (-front.north west) [anchor=south west,
xslant=1, yscale=1/3] {};
\node [cube face] (-side) at (-front.south east) [anchor=south west,
yslant=1, xscale=1/3] {};
}}
}
\starttext
\startsetups[table:initialize]
\setupTABLE[height=2cm,align={middle,lohi}]
\setupTABLE[column][1][[width=4cm,align={right,lohi},frame=on,framecolor=H1prime]
\setupTABLE[column][2][[width=5cm,align={middle,lohi},frame=off]
\setupTABLE[column][3][[width=9cm,align={right,lohi},frame=off]
\setupTABLE[1][1][frame=off]
\stopsetups
\bTABLE[setups=table:initialize]
\switchtobodyfont[10pt]
\bTR
\bTD \eTD
\eTR
\bTR
\bTD $A$ prend la valeur $3$ \eTD
\bTD
%\starttikzpicture
%\pic [cube face/.append style={ball color=green}] at (-1, 1) {cube=3 with
%$A$};\pic [cube face/.append style={ball color=red}] at ( 1, 1) {cube={}
%with $B$};
%\stoptikzpicture
\eTD
\bTD La valeur de $A$ est \color[green]{3} et $B$ n'a pas encore de valeur
\eTD
\eTR
\eTABLE{}
\bTABLE[setups=table:initialize]
\switchtobodyfont[10pt]
\bTR
\bTD \eTD
\eTR
\bTR
\bTD $A$ prend la valeur $3$ \eTD
\bTD
\starttikzpicture
\pic [cube face/.append style={ball color=green}] at (-1, 1) {cube=3 with
$A$};\pic [cube face/.append style={ball color=red}] at ( 1, 1) {cube={}
with $B$};
\stoptikzpicture
\eTD
\bTD La valeur de $A$ est \color[green]{3} et $B$ n'a pas encore de valeur
\eTD
\eTR
\eTABLE{}
\stoptext
The second time
La valeur de $A$ est \color[green]{3}
is green .
--
luigi
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