[pstricks] Retrieving coordinates
Denis Girou
Denis.Girou at idris.fr
Thu May 30 19:06:19 CEST 2002
>>>>> "Arnout.Standaert" == Arnout Standaert <arnout.standaert at agr.kuleuven.ac.be> writes:
Arnout.Standaert> I'm using pstricks to produce A0 posters. Essentially, I put
Arnout.Standaert> textboxes inside a \psshadowbox command and shove these around to get the
Arnout.Standaert> desired layout (I borrowed some code from the textpos package to allow
Arnout.Standaert> absolute positioning of these boxes on the page).
Arnout.Standaert> Now, is there a way of retrieving the coordinates of the lower right corner
Arnout.Standaert> of such a \psshadowbox? This would allow my package to calculate the optimal
Arnout.Standaert> position of new boxes, relative to the already present ones, and save me a
Arnout.Standaert> lot of time shoving the boxes around in a trial-and-error way...
I do not truly understand the question (nor the interest to proceed this
way...) In any case, as you said that you have already some code to compute
what you need, I do not see what is the change with \psshadowbox. You must be
able to consider it as a TeX box like others:
\documentclass[a4paper]{article}
\usepackage{pstricks}
\pagestyle{empty}
\begin{document}
\psframebox[framesep=0,linecolor=red]{\Huge Hello!}
\hfill
\psframebox[framesep=0,linecolor=red]{%
\psshadowbox[shadowsize=0.5,linewidth=1]{\Huge Hello!}}
\end{document}
Anyway, I never build a poster (!), but I would certainely not follow this
direction. In general, computing absolute positions is long, painful, error
proning and absolutely not flexible, ever you use a serie of \rput commands
or of the `textpos' package.
I believe that the "psmatrix" environment is a lot more powerful and
flexible for such needs.
D.G.
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