Node.js start updater and close main program

I am checking if my application has an update by pinging a certain URL, pinging this URL returns whether I need an update or not.

Now, I have a powershell file which actually handles the update, so I'm trying to launch this powershell file from inside of my application.

I have this working, I can spawn my updater file and it will run through and everything is good. However, my application stays open the whole time, which means that once the updater is finished I will have 2 instances of it running.

The obvious solution to this in my mind is to close the application if an update is found (after spawning the updater).

Here is my code:

child = spawn("powershell.exe",['-ExecutionPolicy', 'ByPass', '-File', require("path").resolve(__dirname, '../../../../updater.ps1')]);
child.unref();

self.close();

However, when I try to make the application close, it seems like the updater is never launched. Or rather, I believe it is launched but gets closed when the main application gets closed.

I have the line child.unref() which I thought was supposed to make the spawned window not attached to the main application, but the updater won't stay open.

I have also tried adding {detached: true} as the 3rd parameter of my spawn() command, but it didn't make a difference in the way it was running.

How can I spawn the updater completely separate from my application?

Solution

To start the update separated from your application I think you should use a script instead of a inline parameter. This will ensure that OS creates a separated process from your node app. For example:

var fs = require('fs');
var spawn = require('child_process').spawn;
var out = fs.openSync('./out.log', 'a');
var err = fs.openSync('./out.log', 'a');

var child = spawn('./myscript.sh', [], {
  detached: true,
  stdio: [ 'ignore', out, err ]
});

child.unref();

setTimeout(function(){
    process.exit();
}, 1000);

The myscript.sh looks like this:

sleep 5; ls >> out2.log

The code above will force node exit (after 2 seconds) but just before it started a bash script (which will wait 5 seconds to run ls command). Running this code results in 2 output files (out.log and out2.log). The first one (out.log) is the output of node app calling a child process while the second (out2.log) is the result of script redirected from separated script.

A better approach and more elegant is using on function. But this means that your main process will actually wait for child process to complete the execution. For example:

var fs = require('fs');
var spawn = require('child_process').spawn;
var out = fs.openSync('./out.log', 'a');
var err = fs.openSync('./out.log', 'a');

var child = spawn('ls', [], {
  detached: true,
  stdio: [ 'ignore', out, err ]
});

child.on('exit', (code) => {
    console.log(`Child exited with code ${code}`);
});

child.unref();

In the second example, the ls result will be saved in out.log file. Since the main process will wait for child to complete.

So all depends on what you are willing to achieve. The first solution is not beautiful but will start something really apart from your node app.