如何调用 ShellExecute 使用 C++ 打开具有特定程序的文件而不关联相同的文件扩展名

Question

I read MSDN and tried this line of code

ShellExecute(handle, "open", "fully_qualified_path_to_file", NULL, NULL, SW_SHOWNORMAL);

It works fine, but I need some more functionality. I wrote an app, which makes output in file with .bin extension. On my OS .bin files doesn't associated with any specific program. MSDN says that I can associate all .bin files with my app with programming, but this case is unacceptable for me.

How I can add this line of code to open specific file with specific program? ps I tried to put my application path in 4th argument as shown in MSDN, smith like that,

C:\\\\Application.exe" "%1"

but it didn't work.

Answer 1

If the lpFile parameter is not a .exe then Windows will only be able to execute the file if it has a file type registration in the registry (ShellExecute will read the command line from the registry and replace %1 with the filename).

If you want to force a specific application then lpFile needs to specify the name/path of said application and the file you want it to open has to be part of the parameters in a format supported by the application, usually just the full path to the file (quoted with " if the path contains spaces). ShellExecute will not translate %1 for you in this case.