javascript define arguments

Question

I am trying to create a function in JS that is taking a lot of arguments but does not always use all of them. What I would like to do is define which argument I am putting a refrence in for. This is what I am thinking but not working like I would like it to.

function foo(arg1, arg2, arg3){
let arg1 = arg1;
let arg2 = arg2;
let arg3 = arg3;
}
foo(arg2:'sets value of arg2');

I would like to be able to skip putting in an argument for the first position and only pass info for the second argument. How can I do this?

• Jonas Wilms: Also have a look at named arguments in JavaScript
• dambergn: Thank you to everyone's input, it helped me in making this. github.com/dambergn/js-password-generator

Answer 1

You could spread syntax ... a sparse array with the value for the wanted argument without changing the function's signature.

function foo(arg1, arg2, arg3){
 console.log(arg1, arg2, arg3)
}
foo(...[, 42]);

Or use an object with the key for a specified element

function foo(arg1, arg2, arg3){
 console.log(arg1, arg2, arg3)
}
foo(...Object.assign([], { 1: 42 }));

Answer 2

Try this:

It would require the caller to pass an object.

function foo({arg, arg2}={}) {
 let arg_1 = arg;
 let arg_2 = arg2;
 
 console.log(arg_1);
 console.log(arg_2);
 // Code
}
foo({arg2: 2});

if you need to set the default parameters you can do it this way function foo({arg = '', arg2 = ''}={})

• Fobos: If it helped you, make the right answer to help those in need, but it was a pleasure

Answer 3

You can pass in null as some of the parameters if they are not needed.

function foo(arg1, arg2, arg3){
console.log(arg1, arg2, arg3);
}
foo(null, 30, null);