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最初由 gephen 发布
[B]我想在打开AutoCAD时打开自己指定的文件,应如何编写代码?
我原先想用下面的方法,但会出现指针出错的现象。
void InitAppMsg()
{
AcDbDatabase *pDatabase=new AcDbDatabase(Adesk::kFalse);
pDatabase->rea... [/B]
不用编程,带参数运行ACAD即可

- To start AutoCAD with a command line switch
- Right-click the AutoCAD icon on the Windows Desktop. Then choose Properties.
- In the AutoCAD Properties dialog box, choose the Shortcut tab.
- In the Target box, edit the parameters for the switch using the following syntax:
- "drive:pathname\acad.exe" ["drawing name"] [/switch "name"]
- Valid switches:
- /b Script name (b stands for batch process)
- /t Template file name
- /c Configuration folder
- /v View name
- /s Support folders
- /r Default system pointing device
- /nologo No AutoCAD logo screen
- /p User-defined registry profile
- For example, enter
- "d:\AutoCAD 2002\acad.exe" /t "d:\AutoCAD 2002\template\arch1" /v "plan1" /b "startup"
- Choose OK.
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