Node: spawn a command and print results to a file

Question

If I want to list the files of a directory and save the results in a res.txt I can run the command

ls -l > res.txt

Now I would like to do the same in a process spawned from the main process of Node.

I have tried naively this

spawn(
  'ls',
  ['-l', '>', 'res.txt'],
);

but does not work. Is there any way to achieve what i want?

do you really want to spawn a process, or just get a list of files, inside node? — G-Force, Dec 08 '21 at 23:55

score 0 · Answer 1 · answered Dec 09 '21 at 10:21

0

After some more googling I found that the shell: true option does the trick.

So, if I want a spawned process to run a command which streams its output to a file, I can use this approach

spawn(
  'ls -l > res.txt',
  [],
  {shell: true}
);

answered Dec 09 '21 at 10:21

Picci

score -1 · Answer 2 · answered Dec 09 '21 at 14:25

-1

so from the node documentation, which can be found here:

const { spawn } = require('child_process');
const ls = spawn('ls', ['-lh', '/usr']);

answered Dec 09 '21 at 14:25

G-Force

This is not what the OP asked. This command is to execute a child process and that's it. – Benny Apr 04 '23 at 17:02

2 Answers2