Unable to execute child_process.exec() when path has spaces - node.js

Question

I am using appjs * and I want to execute a command to open a folder.
What I have
var path = __dirname + '/folder to open/';
// path = C:\Program Files\myapplication/folder to open/
require("child_process").exec("start " + path);
Error
Could not find file C:\Program
What I tried
I already tried to escape the spaces, that didn't work.
var path = __dirname + '/folder to open/';
path = path.replace(' ', '\ ');
// path = C:\Program Files\myapplication/folder to open/
require("child_process").exec("start " + path);
When I put the path between quotes, No folder is opened, only another prompt.
var path = "\"" + __dirname + "/folder to open/\"";
path = path.replace(' ', '\ ');
// path = "C:\Program Files\myapplication/folder to open/"
require("child_process").exec("start " + path);
Related bug https://github.com/isaacs/npm/pull/2479
Does anyone has a fix or a workaround?
* link removed

Answer 1

To open a path than contains spaces, you must replace with a double backslash.
In your code you escaped the space character:
"\ "
What you need to do is escape the backslash character so it makes it into the output string:
"\\ "
Try this:
var path = __dirname + '/folder to open/';
// Notice the double-backslashes on this following line
path = path.replace(/ /g, '\\ ');
require("child_process").exec("start " + path);

Answer 2

Well, I fixed it.
Or something like it.
Instead of using
"start " + path
I used
"%SystemRoot%\\explorer.exe \"" + path + "\""
Notice the quotes and the forward slashes.

Answer 3

In my case it is fixed by adding double quotes for the path except the first drive name or letter.
import * as path from 'path'; // npm module
const filePath = 'C:/Program Files/my application/file to open/test.txt';
const rootName = path.parse(filePath).root; // "C:/"
const filePathTo = `${rootName}"${filePath.replace(rootName, '')}"`; // C:/"Program Files/my application/file to open/test.txt"
require("child_process").exec(`start ${filePathTo}`);
The text file will be opened.

Answer 4

this works for me
f= file.replace(/ /g,"\\\ ")

Answer 5

You can also use the old-style 8-character-max/no-space names for each path.
The one I always used was always coding c:\PROGRA~1 instead of c:\Program Files, although this only works on english systems.
If the first 8 characters of any path with more chars are unique, I expect you can do something like newPath = origPath.sub(/\W/g, '').substr(0, 6) + "~1"
Don't have a windows system here, just going by memory.