How to export env variable in node.js

Question

I like my node.js so much, that I want to use it is my bash start up script ~/.bashrc, but I do not know how to export variable.

Currently I have to use this approach:

export PS1=`node ~/PS1.js`
export PS2=`node ~/PS2.js`
export PATH=`node ~/PATH.js`

instead I want .bashrc look have

#!/usr/local/bin/node
//do something, define functions
export_var('PS1', PS1())
export_var('PS2', PS2())
export_var('PATH', generatePATH())

process.env.PATH = something does not export, only sets for the currently executing process, which is node itself.

possible duplicate of How to change value of process.env.PORT in node.js? — rafaelcastrocouto
– rafaelcastrocouto, Commented Oct 20, 2014 at 11:08
try to use "#." as the 1st line and keep the first block of code — rafaelcastrocouto
– rafaelcastrocouto, Commented Oct 20, 2014 at 11:10
@rafaelcastrocouto, you do not understand the question and are trying to close it instead? I need to change the variable from inside node.js and let the parent process keep it. The question you are referring to is about changing PORT variable before the node.js starts. — exebook
– exebook, Commented Oct 20, 2014 at 11:18
You need get the text? You need open process manually inside your main process? Use ' instead ` on .bashrc exports and get the commands as string. — GabrielBiga
– GabrielBiga, Commented Oct 20, 2014 at 11:22
@GabrielBiga, I do not need to get the text of the variable, I need to set it persistently, this is called "export variable" in Linux. — exebook
– exebook, Commented Oct 20, 2014 at 11:27

Community · Accepted Answer · 2017-05-23 11:51:34Z

13

Node.js will run in an separate process which gets a copy of the environment. You cannot change the environment of you parent process (the one executing .bashrc).

But the following question has an answer for you: Can a shell script set environment variables of the calling shell?

You can write a new script file from within node.js and call it via source.

edited May 23, 2017 at 11:51

CommunityBot

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answered Oct 20, 2014 at 11:32

delixfe

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2 Comments

Molten Ice Over a year ago

Could you add a code example for this as I don't understand how I can source a node.js script: "You can write a new script file from within node.js and call it via source."

zdm Over a year ago

How it cal work? Script should be executed in the parent shell, but this is not possible.

Jianwu Chen · Accepted Answer · 2020-12-04 15:27:55Z

12

One possible way is to use JS to print out the export statements, then in shell to use eval to evaluate it in the current shell.

e.g. test.js

#!/usr/bin/env node
console.log('export A=40; export B=10');

In the shell:

eval `./test.js`
echo $A

answered Dec 4, 2020 at 15:27

Jianwu Chen

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1 Comment

bombillazo Over a year ago

This was simple and clean magic! For those using yarn, use yarn --silent <script> when running your command.

Martin Luther ETOUMAN NDAMBWE · Accepted Answer · 2025-11-17 17:27:44Z

0

Create a file named `.env` with your exports.

export MY_VAR="value1"
export MY_OTHER_VAR="value2"

Then launch your app.js

node app.js --env-file=/path/to/.env

Inside your app.js, you just need to look for process.env.MY_VAR and process.env.MY_OTHER_VAR

answered Nov 17 at 17:27

Martin Luther ETOUMAN NDAMBWE

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Comments

zemil · Accepted Answer · 2022-11-01 18:23:12Z

-1

In my case it works:

const exportVar = (name, value) => {
    process.env[name] = value;
};

exportVar("NODE_ENV", "test");

But also in some cases it can be useful to play with exec (or spawn):

const { execSync } = require('child_process');

execSync("NODE_ENV=test && echo $NODE_ENV", {
    stdio: 'inherit',
});

answered Nov 1, 2022 at 18:23

zemil

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