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Color shades responsive to number of divs

I am working on a job board with an always changing quantity of job postings. However each job posting should have a tiered shade of one base color: #219BBE. Here is a sketch of what I am trying to achieve:

Sketch of the shading concept

What I need is a javascript function which changes the shade of a color depending on the number of job_boxes.

So far I found a javascript snippet for generating the shades of #219BBE.

function calculateShades(colorValue) {
  "#219BBE" = colorValue;

// break the hexadecimal color value into R, G, B one-byte components
// and parse into decimal values.
// calculate a decrement value for R, G, and B based on 10% of their
// original values.
var red = parseInt(colorValue.substr(0, 2), 16);
var redDecrement = Math.round(red*0.1);

var green = parseInt(colorValue.substr(2, 2), 16);
var greenDecrement = Math.round(green*0.1);

var blue = parseInt(colorValue.substr(4, 2), 16);
var blueDecrement = Math.round(blue*0.1);

var shadeValues = [];
var redString = null;
var greenString = null;
var blueString = null;

for (var i = 0; i < 10; i++) {
  redString = red.toString(16); // convert red to hexadecimal string
  redString = pad(redString, 2); // pad the string if needed
  greenString = green.toString(16); // convert green to hexadecimal string
  greenString = pad(greenString, 2); // pad the string if needed
  blueString = blue.toString(16); // convert blue to hexadecimal string
  blueString = pad(blueString, 2); // pad the string if needed
  shadeValues[i] = redString + greenString + blueString;

// reduce the shade towards black
  red = red - redDecrement;
  if (red <= 0) {
    red = 0;
  }

  green = green - greenDecrement;
  if (green <= 0) {
    green = 0;
  }

  blue = blue - blueDecrement;
  if (blue <= 0) {
    blue = 0;
  }

}

shadeValues[10] = "000000";
return shadeValues;
}

I simplified the output for this problem: HTML

<!-- Instead of 4 boxes we could also have n boxes -->
<div class="job_box"></div>
<div class="job_box"></div>
<div class="job_box"></div>
<div class="job_box"></div>

CSS

.job_box {
  width: 100%;
  height: 50px;

  /* The dynamic background-color */
  background-color: #219BBE;
}

To count the quantity of job_boxes I would use jQuery:

var numBoxes = $('.job_box').length

And here is the point where I am stuck. I know that I need a loop but that's it. Can you please push me in the right direction?

Fiddle

like image 257
Retador Avatar asked Dec 28 '25 10:12

Retador


2 Answers

That is my solution: DEMO

var n = 0;

$('.job_box').each(function() {
    n++;
    lighten($(this), n);
});

function lighten(that, n) {
    var lightenPercent = 15 / n;
    var rgb = that.css('background-color');
    rgb = rgb.replace('rgb(', '').replace(')', '').split(',');
    var red = $.trim(rgb[0]);
    var green = $.trim(rgb[1]);
    var blue = $.trim(rgb[2]);

    red = parseInt(red * (100 - lightenPercent) / 100);
    green = parseInt(green * (100 - lightenPercent) / 100);
    blue = parseInt(blue * (100 - lightenPercent) / 100);

    rgb = 'rgb(' + red + ', ' + green + ', ' + blue + ')';

    that.css('background-color', rgb);
}

You can, on the other hand, darken your base color by changing the / to * when setting the var for percent.

like image 87
DonJuwe Avatar answered Dec 30 '25 22:12

DonJuwe


Looking at your design, there should only be a limited number of divs (say 4-8) and so I would personally try to keep it simple and pre-calculate the background colours and implement it in 8 lines of CSS rather than loads of JavaScript.

e.g.

DIV.job_box:nth-child(0) { background-color: #219BBE; }
DIV.job_box:nth-child(1) { background-color: <nextcol>; }
DIV.job_box:nth-child(2) { background-color: <nextcol>; }
DIV.job_box:nth-child(3) { background-color: <nextcol>; }
DIV.job_box:nth-child(4) { background-color: <nextcol>; }
DIV.job_box:nth-child(5) { background-color: <nextcol>; }
DIV.job_box:nth-child(6) { background-color: <nextcol>; }

I know this isn't the answer to your specific approach but often we go down paths that are much more complicated than they need to be.

like image 42
Chris Walsh Avatar answered Dec 30 '25 22:12

Chris Walsh