Why doesn't backgroundColor=rgb(a,b,c) work?

Question

Answer 1

You need to use quotes:

document.body.style.backgroundColor = 'rgb(' + a + ',' + b + ',' + c + ')';

JS Fiddle demo.

Or:

document.body.style.backgroundColor = 'rgb(' + [a,b,c].join(',') + ')';

JS Fiddle demo.

Unquoted the JavaScript is passing the variables, as arguments, a,b and c to an undefined function called rgb(). As you're setting a CSS property you need to pass a string, hence the requirement of quoting.

Oh, and also you're using parseInt() which doesn't require a radix to be passed in, but it's better (and easier to avoid problems) if you do (the radix being the expected number-base):

var a = parseInt(prompt("Enter R"), 10) || 255,
    b = parseInt(prompt("Enter G"), 10) || 255,
    c = parseInt(prompt("Enter B"), 10) || 255;

JS Fiddle demo (In the demo I use 105 just so it's clear the default is being used if the cancel button is used).

And if someone hits 'cancel' on the prompt, you might want to supply a default argument to ensure that an actual colour-value is passed, since cancel otherwise, I think, evaluates to false (I'm assuming you'd prefer 255, but obviously adjust to taste).

You could also, of course, simply define a function:

function rgb(r,g,b) {
    return 'rgb(' + [(r||0),(g||0),(b||0)].join(',') + ')';
}
  var a = parseInt(prompt("Enter R"), 10),
      b = parseInt(prompt("Enter G"), 10),
      c = parseInt(prompt("Enter B"), 10);
  document.body.style.backgroundColor = rgb(a,b,c);

JS Fiddle demo

And this approach has the (perhaps specious) benefit of allowing a custom default value to be used:

function rgb(r,g,b, def) {
def = parseInt(def, 10) || 0;
    return 'rgb(' + [(r||def),(g||def),(b||def)].join(',') + ')';
}
var a = parseInt(prompt("Enter R"), 10),
    b = parseInt(prompt("Enter G"), 10),
    c = parseInt(prompt("Enter B"), 10);
document.body.style.backgroundColor = rgb(a,b,c,40);

JS Fiddle demo

References:

• || (logical OR) operator.
• Array.join().
• Element.style.
• parseInt().