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Daniel Victor
Daniel Victor

Reputation: 669

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

<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??

Upvotes: 22

Views: 76415

Answers (7)

Gabriel Pacheco Y&#225;&#241;ez
Gabriel Pacheco Y&#225;&#241;ez

Reputation: 1

This worked for me:

const red = 245;
const green = 128;
const blue = 56;
document.body.style.backgroundColor = `rgb(${red}, ${green}, ${blue})`;

Upvotes: 0

Pluto
Pluto

Reputation: 1

Setting a CSS property to any JavaScript variable requires the value to be passed inside quotation marks (as strings).

document.body.style.backgroundColor = 'red';

check out how we write red inside quotes, as it is a CSS property. Similarly, we write :

document.body.style.backgroundColor = 'rgb(x, y, z)';

Upvotes: 0

David Thomas
David Thomas

Reputation: 253506

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:

Upvotes: 53

malkassem
malkassem

Reputation: 1957

rgb needs to be in quotation mark:

<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>

jsFiddle http://jsfiddle.net/pduQ6/

Upvotes: 1

user554546
user554546

Reputation:

The problem is that the color needs to be a string:

var a = parseInt(prompt("Enter R"),10);
var b = parseInt(prompt("Enter G"),10);
var c = parseInt(prompt("Enter B"),10);
document.body.style.backgroundColor = "rgb(" + [a,b,c].join() + ")";

Demo

Upvotes: 0

MrCode
MrCode

Reputation: 64536

Use quotes around the value

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

Upvotes: 1

Naftali
Naftali

Reputation: 146360

You have no function called rgb(...)

I think you meant to do:

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

Upvotes: 0

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