Using pixel values to define CSS gradient color stops from the end

Question

Using pixel values to define CSS gradient color stops from the end

Currently, I am working on a project involving HTML/CSS/JS where the application has a fixed size and requires precise positioning of elements according to the provided designs. Working with pixel dimensions in CSS is not a concern due to the fixed window size, eliminating the need to worry about browser resizing. Moreover, the focus is solely on ensuring compatibility with webkit and firefox, disregarding IE and Opera.

At certain points in the project, I need to implement a gradient background spanning a specific number of pixels. This can be easily achieved using:

background-image: linear-gradient(to top, #666666, #000000 60px);

(along with its -webkit- and -moz- versions). This method works for most elements but poses a challenge for instances where the color stops need to be defined by top and bottom pixel positions. If these positions were percentages, handling them would be straightforward with code like:

background-image: linear-gradient(to top, #666666, black 60px, transparent 60px, transparent 90%, black 90%, #666666);

(transitioning from grey to black over 60px followed by a transparent section and then black to grey over the last 10%). However, the requirement is to achieve the same outcome based on pixels, considering that the element's size varies. Ideally, relying on JS to dynamically apply the gradient with calculated percentage points should be avoided.

Hence, my inquiry is whether there is a method to specify a color stop x pixels (not percentage) from the end?

css linear-gradients

Answer 1

Answer №1

After conducting a search on the internet, I stumbled upon a fantastic solution proposed by vals using multiple background images. However, instead of utilizing background-size and background-position, I found that incorporating alpha colors (using rgba()) provides more flexibility and stability. Here is an example of how it can be implemented:

background-image:
    /* top gradient - fixed in pixels */
    linear-gradient(to bottom, rgb(128,128,128) 0px, rgba(128,128,128,0) 16px), 
    /* bottom gradient - fixed in pixels */
    linear-gradient(to top, rgb(128,128,128) 0px, rgba(128,128,128,0) 16px),  
    /* background gradient - relative */
    linear-gradient(to bottom, #eee 0%, #ccc 100%);

This approach perfectly met the requirements I had been seeking. :)

Check out the demo here: http://codepen.io/Grilly86/pen/LVBxgQ

Answer 2

After conducting a search on the internet, I stumbled upon a fantastic solution proposed by vals using multiple background images. However, instead of utilizing background-size and background-position, I found that incorporating alpha colors (using rgba()) provides more flexibility and stability. Here is an example of how it can be implemented:

background-image:
    /* top gradient - fixed in pixels */
    linear-gradient(to bottom, rgb(128,128,128) 0px, rgba(128,128,128,0) 16px), 
    /* bottom gradient - fixed in pixels */
    linear-gradient(to top, rgb(128,128,128) 0px, rgba(128,128,128,0) 16px),  
    /* background gradient - relative */
    linear-gradient(to bottom, #eee 0%, #ccc 100%);

This approach perfectly met the requirements I had been seeking. :)

Check out the demo here: http://codepen.io/Grilly86/pen/LVBxgQ

Answer 3

Answer №2

Utilizing the calc() function works perfectly, except for the fact that it doesn't function in MS browsers:

The initial row in each set showcases the solution with two backgrounds stacked, with the second row incorporating the use of calc. Unfortunately, this feature is not supported in Internet Explorer and Edge browsers.

div {
  margin-left: auto;
  margin-right: 0;
  width: 200px;
  height: 20px;
  animation: sweep 5s ease-in-out alternate infinite;
  text-align: center;
  color: white;
  font-family: sans-serif;
  font-weight: bold;
  line-height: 20px;
  will-change: width;
}

div:nth-child(odd) {
  background-image: linear-gradient(to right, red, green 100px, transparent 101px), linear-gradient(to left, red, green 100px);
  border-bottom: 1px solid gray;
}

div:nth-child(even) {
  background-image: linear-gradient(to right, red, green 100px, green calc(100% - 100px), red);
  margin-bottom: 10px;
}

div:nth-child(n+3) {
  width: 300px;
}

div:nth-child(n+5) {
  width: 400px;
}

div:nth-child(n+7) {
  width: 500px;
}

div:nth-child(n+9) {
  width: 600px;
}

@keyframes sweep {
  100% {
    width: 600px;
  }
}

<div> 200 </div>
<div></div>
<div> 300 </div>
<div></div>
<div> 400 </div>
<div></div>
<div> 500 </div>
<div></div>
<div> 600 </div>
<div></div>

Answer 4

Utilizing the calc() function works perfectly, except for the fact that it doesn't function in MS browsers:

The initial row in each set showcases the solution with two backgrounds stacked, with the second row incorporating the use of calc. Unfortunately, this feature is not supported in Internet Explorer and Edge browsers.

div {
  margin-left: auto;
  margin-right: 0;
  width: 200px;
  height: 20px;
  animation: sweep 5s ease-in-out alternate infinite;
  text-align: center;
  color: white;
  font-family: sans-serif;
  font-weight: bold;
  line-height: 20px;
  will-change: width;
}

div:nth-child(odd) {
  background-image: linear-gradient(to right, red, green 100px, transparent 101px), linear-gradient(to left, red, green 100px);
  border-bottom: 1px solid gray;
}

div:nth-child(even) {
  background-image: linear-gradient(to right, red, green 100px, green calc(100% - 100px), red);
  margin-bottom: 10px;
}

div:nth-child(n+3) {
  width: 300px;
}

div:nth-child(n+5) {
  width: 400px;
}

div:nth-child(n+7) {
  width: 500px;
}

div:nth-child(n+9) {
  width: 600px;
}

@keyframes sweep {
  100% {
    width: 600px;
  }
}

<div> 200 </div>
<div></div>
<div> 300 </div>
<div></div>
<div> 400 </div>
<div></div>
<div> 500 </div>
<div></div>
<div> 600 </div>
<div></div>

Answer 5

Answer №3

It might not seem feasible, but stacking two elements where one has opaque pixels at the bottom and the other has pixels at the top could be achieved without the need for JavaScript.

.background {
    position: absolute;
    background-image: linear-gradient(to top, #666666, black 60px, transparent 60px);
}
.overlay {
    position: relative;
    background-image: linear-gradient(to bottom, #666666, black 60px, transparent 60px);
}

Answer 6

It might not seem feasible, but stacking two elements where one has opaque pixels at the bottom and the other has pixels at the top could be achieved without the need for JavaScript.

.background {
    position: absolute;
    background-image: linear-gradient(to top, #666666, black 60px, transparent 60px);
}
.overlay {
    position: relative;
    background-image: linear-gradient(to bottom, #666666, black 60px, transparent 60px);
}

Answer 7

Answer №4

Continuing from the previous response by po228, we can create two different gradients within the same element.

Specify one gradient to start from the top and the other from the bottom.

.test {
  background: linear-gradient(to top, red 10px, white 10px),
      linear-gradient(to bottom, blue 10px, white 10px);
  background-size: 100% 50%;
  background-repeat: no-repeat;
  background-position: bottom center, top center;
  height: 150px;
  width: 300px;
}

<div class="test"></div>

Answer 8

Continuing from the previous response by po228, we can create two different gradients within the same element.

Specify one gradient to start from the top and the other from the bottom.

.test {
  background: linear-gradient(to top, red 10px, white 10px),
      linear-gradient(to bottom, blue 10px, white 10px);
  background-size: 100% 50%;
  background-repeat: no-repeat;
  background-position: bottom center, top center;
  height: 150px;
  width: 300px;
}

<div class="test"></div>

Using pixel values to define CSS gradient color stops from the end

Answer №1

Answer №2

Answer №3

Answer №4

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