我的代码只是针对最初的问题设计的——它不是通用的。 这将以其中心处的基点缩放正方形。
(defun c:scl(/ ss scl st 3p)
(vl-load-com)
(or scale:def (setq scale:def 0.)
(if (setq ss (ssget
(list (cons 0 "*Polyline")
(cons 8 "Columns")
(if (getvar "CTAB")
(cons 410 (getvar "CTAB"))
(cons 67 (- 1 (getvar "TILEMODE")))))))
(progn
(initget 6)
(setq scl (getreal (strcat "\nSpecify Scale Factor <" (rtos scale:def) "> : ")))
(or (not scl) (setq scale:def scl))
(foreach x(mapcar 'vlax-ename->vla-object
(mapcar 'cadr (ssnamex ss)))
(setq st (vlax-curve-getStartPoint x)
3p (vlax-curve-getPointatParam x 2.0))
(vla-scaleEntity x
(vlax-3D-point
(polar st (angle st 3p) (/ (distance st 3p) 2.0))) scale:def)))
(princ "\n<!> No Lines Found <!>"))
(princ))
尽管如此，它还是做出了很多假设——比如它们是完全正方形的，等等
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