How to open Explorer with a specific file selected?

Question

I would like to code a function to which you can pass a file path, for example:

C:\\FOLDER\\SUBFOLDER\\FILE.TXT

and it would open Windows E

Answer 1

Easiest way without using Win32 shell functions is to simply launch explorer.exe with the /select parameter. For example, launching the process

explorer.exe /select,"C:\Folder\subfolder\file.txt"

will open a new explorer window to C:\Folder\subfolder with file.txt selected.

If you wish to do it programmatically without launching a new process, you'll need to use the shell function SHOpenFolderAndSelectItems, which is what the /select command to explorer.exe will use internally. Note that this requires the use of PIDLs, and can be a real PITA if you are not familiar with how the shell APIs work.

Here's a complete, programmatic implementation of the /select approach, with path cleanup thanks to suggestions from @Bhushan and @tehDorf:

public bool ExploreFile(string filePath) {
    if (!System.IO.File.Exists(filePath)) {
        return false;
    }
    //Clean up file path so it can be navigated OK
    filePath = System.IO.Path.GetFullPath(filePath);
    System.Diagnostics.Process.Start("explorer.exe", string.Format("/select,\"{0}\"", filePath));
    return true;
}

Reference: Explorer.exe Command-line switches