请高手帮忙修改批量框剪线程序

mycad

;;此程序来自mjtd.com
;;本人lisp不是太精通，请高手把下面的程序改为批量剪切“框内”的线 ，先谢谢了！
(defun c:mt(); / e1 pt x ptx pty l1 i p1 p2 p1x p1y point count  rect e0
(setvar "osmode" 0)
(setq l1 nil)
(setq i 0)
(setq rect (car (entsel "\n请选择需剪切的矩形:")))
(setq e0 (entget rect))
(while (setq x (nth i e0))
(if (= (car x) 10)
(progn
(setq ptx (nth 1 x))
(setq pty (nth 2 x))
(setq x (list ptx pty))
(setq l1 (cons x l1))
)
)
(setq i (1+ i))
)
(reverse l1)
(setq p1 (car l1))
(setq p1x (nth 0 p1))
(setq p1y (nth 1 p1))
(setq count 0)
(repeat 3
(setq count (+ count 1))
(setq pt (nth count l1))
(setq ptx (nth 0 pt))
(setq pty (nth 1 pt))
(setq point (mapcar '+ p1 pt))
(setq point (mapcar '/ point '(-1 -1 -1)))
(if
(and (/= (nth 0 point) p1x)
(/= (nth 0 point) ptx)
(/= (nth 1 point) p1y)
(/= (nth 1 point) pty)
)
(setq p point)
)
)
(setq l1 nil)
(command "offset" 0.001 rect p "")
(setq e0 (entlast))
(setq e1 (entget e0))
(princ e1)
(setq i 0)
(while (setq x (nth i e1))
(if (= (car x) 10)
(progn
(setq ptx (nth 1 x))
(setq pty (nth 2 x))
(setq x (list ptx pty))
(setq l1 (cons x l1))
)
)
(setq i (1+ i))
)
(reverse l1)
(setq p1 (car l1))
(setq p2 p1)
(command "erase" e0 "")
(setq count 0)
(repeat 3
(setq count (+ count 1))
(setq pt (nth count l1))
(command "trim" rect "" "f" p1 pt "" "")
(setq p1 pt)
)
(command "trim" rect "" "f" p1 p2 "" "")
(setvar "osmode" 687)
)

mycad

;;为什么这样就可以了？
(defun c:mt1(); / e1 pt x ptx pty l1 i p1 p2 p1x p1y point count  rect e0
(setvar "osmode" 0)
(setq l1 nil)
(setq i 0)
(setq rect (car (entsel "\n请选择需剪切的矩形:")))
(setq e0 (entget rect))
(while (setq x (nth i e0))
(if (= (car x) 10)
(progn
(setq ptx (nth 1 x))
(setq pty (nth 2 x))
(setq x (list ptx pty))
(setq l1 (cons x l1))
)
)
(setq i (1+ i))
)
(reverse l1)
(setq p1 (car l1))
(setq p1x (nth 0 p1))
(setq p1y (nth 1 p1))
(setq count 0)
(repeat 3
(setq count (+ count 1))
(setq pt (nth count l1))
(setq ptx (nth 0 pt))
(setq pty (nth 1 pt))
(setq point (mapcar '+ p1 pt))
(setq point (mapcar '/ point '(2.0 2.0 2.0)))
(if
(and (/= (nth 0 point) p1x)
(/= (nth 0 point) ptx)
(/= (nth 1 point) p1y)
(/= (nth 1 point) pty)
)
(setq p point)
)
)
(command "undo" "begin" "")
(setq l1 nil)
(command "offset" 0.001 rect p "")
(setq e0 (entlast))
(setq e1 (entget e0))
(princ e1)
(setq i 0)
(while (setq x (nth i e1))
(if (= (car x) 10)
(progn
(setq ptx (nth 1 x))
(setq pty (nth 2 x))
(setq x (list ptx pty))
(setq l1 (cons x l1))
)
)
(setq i (1+ i))
)
(reverse l1)
(setq p1 (car l1))
(setq p2 p1)
(command "erase" e0 "")
(setq count 0)
(repeat 3
(setq count (+ count 1))
(setq pt (nth count l1))
(command "trim" rect "" "f" p1 pt "" "")
(setq p1 pt)
)
(command "trim" rect "" "f" p1 p2 "" "")
  (command "undo" "end" "")
(setvar "osmode" 687)
)

zyhandw

真的能行？我是没得到正确结果！p1、p2为同一点，能trim？