如何以编程方式打开控制面板? [英] How to programmatically open control panel?
问题描述
如何以编程方式打开自定义控制面板,例如custom.cpl?具体来说,当以32位应用程序运行时,如何打开64位cpl?
Vista添加了对规范名称的支持,因此您不必对dll文件名和制表符索引进行硬编码
示例: WinExec(%systemroot%\ system32 \ control.exe/name Microsoft.WindowsUpdate",SW_NORMAL);
(名称始终为英文)
有关列表,请参见 MSDN >
XP/2000支持"control.exe鼠标"和一些其他关键字,请参见同一MSDN页面以获取列表(您可以通过在control.exe上运行字符串来找到一些未记录的内容)
How do I open a custom control panel programmatically, like custom.cpl? Specifically, how do I open a 64-bit cpl when running as 32-bit application?
Vista added support for canonical names so you don't have to hard code dll filenames and tab indexs
Example: WinExec("%systemroot%\system32\control.exe /name Microsoft.WindowsUpdate", SW_NORMAL);
(Names are always in english)
See MSDN for a list
XP/2000 supports "control.exe mouse" and a few other keywords, see the same MSDN page for a list (You can probably find some undocumented ones by running strings on control.exe)
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