NodeJS: How can I include some javascript code snippet without using require?

Question

I wanted to include some js files without using require because it would link it to a variable, and i have some functions i would like to call directly. how can i do it ? is it bad practice ?

what i want to avoid is this: let's say i have tool.js as follow:

function foo() {
    log.debug("foo");
}    
exports.foo = foo;

in app.js

var tool= require('tools.js');

tool.foo();

i would like to be able to call foo without creating a module for it as if it was define in app.js; like so

foo();

You can find the answer here (Please check the third answer): stackoverflow.com/questions/5797852/… — Ami Hollander
– Ami Hollander, Commented Jun 18, 2016 at 15:20
Yes the third answer is right! sorry, i just checked the 2 first. Thanks! — Xsmael
– Xsmael, Commented Jun 18, 2016 at 15:34

Gabs00 · Accepted Answer · 2016-06-18 15:25:53Z

2

You can require a file and use its functions without assigning it to a variable by using the global object.

file1.js

function logger(){
    console.log(arguments);
}

global.logger = logger;

file2.js

require('./file1');

logger('ABC');

This approach would get rid of variable scoping and would pollute the global namespace potentially leading to clashes with variable naming.

answered Jun 18, 2016 at 15:25

Gabs00

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3 Comments

Xsmael Over a year ago

would this work if global.logger = logger; was missing in file1.js ? coz i saw someone doing it in a tutorial but it wasn"t working for me thanks for the warning by the way

Gabs00 Over a year ago

It would not work, tested that first and the function was not available @Xsmael

Xsmael Over a year ago

This is very weird! now it works for me too, and i don't know why -_-

Alongkorn · Accepted Answer · 2016-06-19 04:24:27Z

0

You need to use global like this,

======= app.js ======

global.foo= require('tools.js'); // declare as global
foo(); // can be called from all files

answered Jun 19, 2016 at 4:24

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Alongkorn

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