如果 ShellExecute 由于 C++ 中没有文件关联而失败，如何打开窗口的默认对话框? [英] How to open a default dialog for window if ShellExecute fails due to no file association in C++?

问题描述

只要文件关联正确，我就可以使用 Windows ShellExecute 函数打开文件而不会出现问题.

I can use the windows ShellExecute function to open a file with no problems so long as the file has a correct association.

如果不存在关联，我想使用默认窗口对话框打开文件:

If no association exists i would like to use the default windows dialog to open the file:

这可能吗?如果是这样怎么办?

Is this possible? If so how?

推荐答案

记录 方式该对话框使用 openas 动词.

The documented way to show that dialog is to use the openas verb.

CoInitializeEx(NULL, COINIT_APARTMENTTHREADED|COINIT_DISABLE_OLE1DDE);
SHELLEXECUTEINFO sei = { sizeof(sei) };
sei.fMask = SEE_MASK_NOASYNC;
sei.nShow = SW_SHOWNORMAL;
sei.lpVerb = "openas";
sei.lpFile = "C:\\yourfile.ext";
ShellExecuteEx(&sei);

如果您在 HKEY_CLASSES_ROOT\Unknown\shell\openas 下检查，您会发现这与在 shell32 中调用(未记录的)OpenAs_RunDLL 导出相同.

If you check under HKEY_CLASSES_ROOT\Unknown\shell\openas you see that this is the same as calling the (undocumented) OpenAs_RunDLL export in shell32.

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