Open a directory in File Explorer - node.js

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

Would be good to use this package so it would open on diffrent platform
https://www.npmjs.com/package/open-file-explorer
Or just use this part of it
function openExplorerin(path, callback) {
var cmd = ``;
switch (require(`os`).platform().toLowerCase().replace(/[0-9]/g, ``).replace(`darwin`, `macos`)) {
case `win`:
path = path || '=';
cmd = `explorer`;
break;
case `linux`:
path = path || '/';
cmd = `xdg-open`;
break;
case `macos`:
path = path || '/';
cmd = `open`;
break;
}
let p = require(`child_process`).spawn(cmd, [path]);
p.on('error', (err) => {
p.kill();
return callback(err);
});
}

Answer 3

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.

Answer 4

A slightly simpler, and more cross-platform solution, would be to use this.
var explorer;
switch (platform()) {
case "win32": explorer = "explorer"; break;
case "linux": explorer = "xdg-open"; break;
case "darwin": explorer = "open"; break;
}
spawn(explorer, [path], { detached: true }).unref();
platform() is from the os module, and spawn is from the child_process module.