How to open a default dialog for window if ShellExecute fails due to no file association in C++?

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.

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

Is this possible? If so how?

回答1:

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);

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.

回答2:

Execute RUNDLL32 Shell32,OpenAs_RunDLL path/to/file/to/open

回答3:

Simply do not use explicit verb. Using a specific verb like 'open' is a big mistake:

'open' may be not a default verb (for example, it may be 'play', 'edit' or 'run')
'open' may not exists

It is a way more correct to simply pass NULL as verb. The system will automatically select most appropriate verb:

Default verb will be used, if it is set
'open' verb will be used, if no default verb is set
first verb will be used, if no default and 'open' verbs are available
if no verbs are assigned - the system will bring "Open with" dialog

In other words, simple

ShellExecute(0, NULL, 'C:\MyFile.StrangeExt', ...);

will show "Open with" dialog.

Only use a specific verb if you want a specific action. E.g. 'print', 'explore', 'runas'. Otherwise - just pass nil.

来源：https://stackoverflow.com/questions/6364879/how-to-open-a-default-dialog-for-window-if-shellexecute-fails-due-to-no-file-ass

标签

winapi

shellexecute

fileopendialog