I've got a parent application in node.js which needs to spawn multiple worker applications (also in node.js) applications according to need.
I've already got communication working between them - don't need to use any of the built-in node stuff.
Now the problem is that I'd like each worker process to have it's own console window - since I do a lot of writing to the console and I want to keep an eye on it.
I've looked through the Node child_process documentation, and it says that by setting options to detached:
On Windows, setting options.detached to true makes it possible for the child process to continue running after the parent exits. The child will have its own console window.
However when I use my own code
const Process = require("child_process").spawn;
Process(process.argv[0], ["myApplicationPath","otherArgs"],{detached: true,stdio: ['ignore']});
It doesn't work. The child application does spawn, but no console window turns up.
I'm a bit late here, but I just had to figure this out as well, so here is the answer for anyone else who is struggling with this:
I managed to spawn my child application in its own console using this:
childProcess.spawn("<cmd>", [], {shell: true, detached: true});
In addition to the {detached: true} what OP is using, I used {shell: true}. With the combination of both, I managed to spawn my child application with its own console.
Related
The following Node.js script can restart itself and will even still print to the correct console (or terminal if you prefer), but it will no longer be running in the foreground, as in you can't exit it with Ctrl+C anymore (see screenshot) etc:
console.log("This is pid " + process.pid);
setTimeout(function () {
process.on("exit", function () {
require("child_process").spawn(process.argv.shift(), process.argv, {
cwd: process.cwd(),
detached : true,
stdio: "inherit"
});
});
process.exit();
}, 5000);
I've already tried detached: true vs detached: false, but obviously this didn't solve the problem...
Is there a way to make the new node process run in the foreground, replacing the old one? Or this this not possible?
I know that in Bash you can pull a program back from the background like this:
$ watch echo "runs in background" &
$ fg # pulls the background process to the foreground
But I'm not looking for a Bash command or so, I'm looking for a programmatic solution within the Node.js script that works on any platform.
No, once a process has exited it cannot perform any more operations, and there's no such thing as a "please foreground this after I exit"-type API for terminals that I've ever heard of.
The proper way to solve this is via a wrapper which monitors your process for failures and restarts. The wrapper then has control of stdio and passes those to its children.
You could achieve this via a simple bash loop, another node script, or you might just be able to leverage the NodeJS cluster module for this.
As Jonny said, you'd need to have a process manager that handles the running of your application. Per the Node.js documentation for child_process, spawn() functions similar to popen() at the system level, which creates a forked process. This generally doesn't go into the foreground. Also, when a parent process exits, control is returned to either the calling process or the shell itself.
A popular process management solution is PM2, which can be installed via npm i -g pm2. (Not linking to their site/documentation here) PM2 is cross-platform, but it does require an "external" dependency that doesn't live within the codebase itself.
I would also be curious as to why you want a script that on exit restarts itself in the manner you're describing, since it seems like just re-running the script -- which is what PM2 and similar do -- would yield the same results, and not involve mucking around with process management manually.
I am using child_process.spawn() to start a script from my Node.JS application running on Ubuntu. As far as I know, standard forked or spawned *nix processes don't generally die when the parent dies, but when spawning processes from Node.JS, they seem to get killed when my application crashes, or is aborted with ctrl-c etc.
Why is this and is there a way around this? I can't seem to find any obvious option in the child_process API.
My application starts some quite long-running tasks that should run in the background, and if my node server crashes or is restarted for some other reason I don't want to interrupt the tasks, instead I want the node server to come back up and gracefully resume monitoring the progress of those running tasks.
you need to set the detached option
If the detached option is set, the child process will be made the
leader of a new process group. This makes it possible for the child to
continue running after the parent exits.
var child = spawn('prg', [], {
detached: true,
stdio: [ 'ignore', out, err ]
});
How can I launch a user-defined application from an electron application. After the application is launched the electron application should quit, and restart after the program has finished. This to free system resources.
I was hoping to be able to use a child_process.spawn with detach option and unref.
let command= 'notepad.exe && ./electron.exe'
const launched_application = child_process.spawn(command,[],{detached:true,stdio:'ignore',shell:true})
launched_application.unref()
app.quit()
When the code hits app.close() the child_process is also killed. Is there a way to keep the child_process running after app.quit()
This also gives an arror on the && part, which I'm planning to fix using a .bat file. Any recommendations regarding this are also welcome.
if you want to relaunch app again then you should just close or hide all windows and activate setInterval to keep checking if child process is alive or not with help of pid ...
after that you can create new window again or show hidden window... and if you want fresh process then you should relaunch app after you get callback of user quitting that child application....
I am spawning a child process with node, which is an external application on OSX. The problem is that this external application appears BEHIND the terminal window. I want this application to appear in the front, on top.
My code looks something like this:
var spawn = require('child_process').spawn,
process = spawn('myApp');
The application launches, but it does not show up to the user because it's behind the terminal window. I want it to be in front.
I am using child_process.spawn() to start a script from my Node.JS application running on Ubuntu. As far as I know, standard forked or spawned *nix processes don't generally die when the parent dies, but when spawning processes from Node.JS, they seem to get killed when my application crashes, or is aborted with ctrl-c etc.
Why is this and is there a way around this? I can't seem to find any obvious option in the child_process API.
My application starts some quite long-running tasks that should run in the background, and if my node server crashes or is restarted for some other reason I don't want to interrupt the tasks, instead I want the node server to come back up and gracefully resume monitoring the progress of those running tasks.
you need to set the detached option
If the detached option is set, the child process will be made the
leader of a new process group. This makes it possible for the child to
continue running after the parent exits.
var child = spawn('prg', [], {
detached: true,
stdio: [ 'ignore', out, err ]
});