Open a directory in File Explorer

Question

On ms Windows from node.js code, how can I open a specific directory (ex: c:\documents) in Windows file explorer?

I guess in c#, it would be:

process.Start(@"c:\test")

Answer 1

Try the following, which opens a File Explorer window on the computer running Node.js:

require('child_process').exec('start "" "c:\\test"');

If your path doesn't contain whitespace, you can also get away with 'start c:\\test', but the above - which requires "" as the 2nd argument^[1] is the most robust approach.

Note:

• The File Explorer window will launch asynchronously, and will receive focus when it does.

• This related question asks for a solution that prevents the window from "stealing" focus.

---

^{[1] cmd.exe's internal start command by default interprets a "..."-enclosed 1st argument as the window title for the new console window to create (which doesn't apply here). By supplying a (dummy) window title - "" - explicitly, the 2nd argument is reliably interpreted as the target executable / document path.}

Answer 2

I found another way on the WSL and potentially windows itself.

Note that you have to make sure you're formatting the path for Windows not Linux (WSL). I wanted to save something on Windows, so in order to do that you use /mnt directory on WSL.

// format the path, so Windows isn't mad at us
// first we specify that we want the path to be compatible with windows style
// then we replace the /mnt/c/ with the format that windows explorer accepts
// the path would look like `c:\\Users\some\folder` after this line
const winPath = path.win32.resolve(dir).replace('\\mnt\\c\\', 'c:\\\\');

// then we use the same logic as the previous answer but change it up a bit
// do remember about the "" if you have spaces in your name
require('child_process').exec(`explorer.exe "${winPath}"`);

This should open the file explorer for you.