Executing shell command using child process

Question

I am trying to execute a command to fork a chain using ganache. I have struggled to find a way to do the fork programmatically so I want to try execute it through a child process.

The code I am using is:

const { exec } = require("child_process");
const ganache = require("ganache");

const ls = exec('ganache-cli --fork https://speedy-nodes-nyc.moralis.io/KEY_HERE/bsc/mainnet/archive', function (error, stdout, stderr) {
  if (error) {
    console.log(error.stack);
    console.log('Error code: ' + error.code);
    console.log('Signal received: ' + error.signal);
  }
  console.log('Child Process STDOUT: ' + stdout);
  console.log('Child Process STDERR: ' + stderr);
});

ls.on('exit', function (code) {
  console.log('Child process exited with exit code ' + code);
});

But when I run this, No output is given at all. No errors or anything it just carries on with the program asif this code did not exist. When I replace the command with 'dir', it works perfectly and lists the files in the directory. Even if I remove the node address and just use "ganache-cli --fork" as the command, nothing happens.

Why does nothing happen when I add the ganache commands?

Answer 1

If you take a look at the child process docs, they say exec:

spawns a shell and runs a command within that shell, passing the stdout and stderr to a callback function when complete.

However, Ganache is a process that continues running and doesn't "complete" until you kill it. This allows you to send multiple requests to Ganache without it shutting down on you.