How to make a node.js server run with a specific command

Question

I want to start my index.js file with node --max-old-space-size=1024 index.js but I don't want to type the command manually everytime, any Idea how can I make a .js file that executes the command for me? Like start.js and when I launch it, it immediately runs node --max-old-space-size=1024 index.js. This might be super straight forward but I'm still new to node.js 😅

You could spawn your "worker" node process, from a "starter" node process if you dont want so use a package scribt as J.F. answered. Use for that the node "child_process" module. — Marc
– Marc, Commented Oct 28, 2021 at 18:15

J.F. · Accepted Answer · 2021-10-28 17:05:05Z

3

You can modify your package.json to create a command that execute node --max-old-space-size=1024 index.js

In your package.json

"scripts": {
  "start": "node --max-old-space-size=1024 index.js"
}

And then run npm run start.

answered Oct 28, 2021 at 17:05

J.F.

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3 Comments

Karim Walid Over a year ago

Can't I make it an independent .js file?

Karim Walid Over a year ago

Like without using command line just click it and it executes the other command right away

J.F. Over a year ago

As far as I know, if you want to add parameters --max-old-space-size=1024 you will need to add into the call for the other JS file. You can do a file start.js where exists the line require('./index.js') but your parameter (--max-old-space-size=1024) will be necessary to run start.js too. So is a better way use package.json I think.

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