How to correctly build js using gulp, taking into account environment variables?

Question

JavaScript

There is a frontend on Angular1. There is a service with a method in which the URL of the API server is hardcoded. How to build JS using gulp so that if the variable ENV == 'develop', then the server address changes to another one.
Of course, I can stupidly replace the regex, but I want to do it right.
For example, to compile pug, I use gulp-data to pass json with variables to the compiler. How to do it with JS?

Answer 1

Maybe the stars didn't line up that way, but my envify didn't take off. Solved the problem like this:

// /gulpfile.babel.js
fs.writeFileSync('./.env.json', JSON.stringify(process.env, ["NODE_ENV"]));

// /src/js/modules/App.API/factoryes/API.factory.js
(function () {
    'use strict';

        window.process = {env: require('../../.env.json')};

        if (process.env.NODE_ENV === "development") {
            self.url = "https://api.domain.dev";
        }
        else {
            self.url = "https://api.domain.ru";
        }

})();

Answer 2

Watch what you collect. For webpack EnvironmentPlugin , for browserify -- envify , or even easier to get out with aliases.