I've been able to successfully run commands using the exec() command. However, I'd like to leave a process running and continue to run commands on the open process, then close on app exit. Take this generic code:
const { exec } = require("child_process");
exec("XR_Command -i 192.168.0.100 -f /ch/01/on | kill", (error, stdout, stderr) => {
if (error) {
console.log(`error: ${error.message}`);
return;
}
if (stderr) {
console.log(`stderr: ${stderr}`);
return;
}
console.log(`stdout: ${stdout}`);
});
I would like to keep the XR_Command process active so that I can issue commands to the active process. So, basically I would like to do this:
> XR_Command -i 192.168.0.100
> /ch/01/on
> /ch/02/on
> /ch/03/on
I cannot for the life of me figure out how to make this function properly by referencing the existing child process. Thanks!
child_process.spawn()instead, so you can set thestdioconfiguration so you get a handle to the child process's input handle.