Home/ Questions/Q 9175383
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-17T16:54:16+00:00

<html> <head> <title> Colors </title> </head> <body> <script type=text/javascript> var a = parseInt(prompt(Enter R));

  • 0
<html>
    <head>
        <title> Colors </title>
    </head>

    <body>
    <script type="text/javascript">
        var a = parseInt(prompt("Enter R"));
        var b = parseInt(prompt("Enter G"));
        var c = parseInt(prompt("Enter B"));
        document.body.style.backgroundColor=rgb(a,b,c);
    </script>
    </body>
</html>

Why doesn’t the background color change according to the RGB values? What have I done wrong??

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    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:

    • 0