How to find if an document can be OPENed via ShellExecute?

问题

I want to check if a particular file can be successfully "OPEN"ed via ShellExecute, so I'm attempting to use AssocQueryString to discover this.

Example:

DWORD size = 1024;
TCHAR buff[1024];  // fixed size as dirty hack for testing

int err = AssocQueryString(0, ASSOCSTR_EXECUTABLE, ".mxf", NULL ,buff , &size);
openAction->Enabled  = ((err == S_OK) || (err == S_FALSE)) && (size > 0);

Now, this almost works. If there's a registered application, I get the string.

But, there's a catch: On Vista, even if there is no registered application, It returns that the app c:\Windows\System32\shell32.dll is associated, which is the thing that brings up the 100% useless "Windows cannot open this file: Use the Web service to find the correct program?" dialog.

Obviously I want to hide that peice of cr*p from end users, but simply comparing the returned string to a constant seems like an ugly, brute-force and fragile way of doing it.

Also, hacking the registry to totally disable this dialog isn't a great idea.

What's a better option?

回答1:

I always use FindExecutable() to get the registered application for a given document.

回答2:

There is another way to do this, using the ASSOCF_INIT_IGNOREUNKNOWN option flag with AssocQueryString().

int err = AssocQueryString(ASSOCF_INIT_IGNOREUNKNOWN, ASSOCSTR_EXECUTABLE, ".mxf", NULL ,buff , &size);

This has a couple of important advantages over using FindExecutable()

It can work with just the file extension, while FindExecutable needs a full path to an existing file of the specified type.
Because it's not accessing the file, it's much faster with Samba and other network storage. Calling FindExecutable() on one file in a directory containing ~3000 files via Samba took > 1 second in my tests.

来源：https://stackoverflow.com/questions/1265099/how-to-find-if-an-document-can-be-opened-via-shellexecute

标签

c++

winapi

shellexecute