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?

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 = ''}={})

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);