将lisp变大以获得矩形

CadFrank · 发表于 2022-7-6 09:38:35

这是维度块

paulmcz · 发表于 2022-7-6 09:44:20

由于没有人知道块的尺寸，下面的代码适合插入第一个块。对于第二个，需要计算第二个插入点的位置。

(command "_.insert" "path or block's name" point2 "x-scale" "y-scale" "rotation")

如果块已在图形中，请跳过路径

paulmcz · 发表于 2022-7-6 09:46:32

好的，第二个块必须具有插入点=p3=>

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(setq p3 (polar point2 (* pi 1.5) 38.0))

CadFrank · 发表于 2022-7-6 09:50:32

假设这是我开始的代码

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[font=Courier New](defun c:BOX ( / pt1 pt2 pt3 pt4 )(setq pt1 (getpoint "Pick first corner: "))(setq pt3 (getcorner pt1 "Pick opposite corner: "))(setq pt2 (list (car pt3) (cadr pt1)))(setq pt4 (list (car pt1) (cadr pt3)))(command "line" pt1 pt2 pt3 pt4 "c" )[/font](command "_.insert" "path or block's name" point2 "x-scale" "y-scale" "rotation")(setq p3 (polar point2 (* pi 1.5) 38.0))))

这样可以吗！

paulmcz · 发表于 2022-7-6 09:51:47

如果块的名称为“123”，则它位于图形中，并且您希望接受x比例为1.0、y比例为1.0和旋转为0.0。

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(defun c:BOXX (/ osn pt1 pt2 p3)
 (setq osn (getvar "osmode"))
 (setvar "osmode" 0)
 (setq pt1 (getpoint "Pick first corner: "))
 (setq pt2 (getcorner pt1 "Pick opposite corner: "))
 (command "rectang" pt1 pt2)
 (command "_.insert" "123" pt2 "" "" "" )
 (setq p3 (polar pt2 (* pi 1.5) 38.0))
 (command "_.insert" "123" p3 "" "" "")
 (setvar "osmode" osn)
 (princ)
)