Animated CSS background pattern, sliding infinitely

问题

I am trying to create an animated repeating-pattern (diagonal stripes sliding horizontally), as placeholder for a loading block (li in this case).

How can I make the animation smooth/continuous giving the illusion the pattern is infinitely sliding?

How to calculate the element width so that the pattern is continuous? (stripes shouldn't look broken/interrupted).
How to make it loop looking like it is not restarting but rather sliding infinitely? (the 100% frame should pass to the 0% frame without any glitch)

The goal is to have a class I can add to any block and will visually look like loading/processing.

Note: no JS; pure CSS.

li {
  display: inline-block;
  width: calc(20px * 8); /* how to calculate this, relative to the width (of the pattern or the step), to achieve pattern continuity exactly?
    Of course without doing trying&error to know it should be 24.75px * 8.
  */
  height: 200px;
  background-color: blue;

  background-image: repeating-linear-gradient(-45deg, transparent, transparent 10px, black 10px, black 20px);
  animation: loading-slide 1s linear infinite;

}
  @keyframes loading-slide {
    from { background-position: 0% 0% }
    to { background-position: 100% 0% }
  }

<ul>
    <li>test
    <li>test
</ul>

回答1:

The correct formula should be (20px / cos(45deg)) * N. Then you can make the background-size to be 200% 100% (twice bigger than the element) and you animate it from left to right:

li {
  display: inline-block;
  width: calc( (20px / 0.707) * 3); /*cos(45deg) = 0.707*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(-45deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

.alt li {
  width: calc( (20px / 0.707) * 6);
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

You can consider any degree and adjust the formula as needed. (20px / cos(90deg - |Xdeg|)) * N with X between -90deg and 90deg

Example with -60deg

li {
  display: inline-block;
  width: calc((20px / 0.866) * var(--n,3)); /*cos(30deg) = 0.866*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(-60deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

.alt li {
  --n:6;
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

Example with 30deg

li {
  display: inline-block;
  width: calc((20px / 0.5) * var(--n,8)); /*cos(60deg) = 0.5*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(30deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

.alt li {
  --n:12;
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

Example with 80deg

li {
  display: inline-block;
  width: calc((20px / 0.9848) * var(--n,8)); /*cos(10deg) = 0.9848*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(80deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

.alt li {
  --n:12;
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

You can clearly identify the trivial case where X=+/-90deg (vertical stripes) and we will have cos(0)=1 thus the formula will be 20px * N. Also when X=0 (horizontal stripes) we will have cos(90deg) = 0 and any width will work since there is no vertical pattern (the formula is no more defined)

li {
  display: inline-block;
  width: calc(20px * var(--n,8)); 
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(90deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

.alt li {
 background-image:repeating-linear-gradient(0deg, transparent, transparent 10px, black 10px, black 20px);
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

What about value outside [-90deg,90deg]?

The above range already cover 180deg and since we are dealing with something symetric, all the value can be represented inside that range.

Example: 110deg is the same as -70deg

li {
  display: inline-block;
  width: calc((20px / 0.9396) * var(--n,8)); /*cos(20deg) = 0.9396*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(110deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}
.alt li {
  --n:12;
  background-image: repeating-linear-gradient(-70deg, transparent, transparent 10px, black 10px, black 20px);
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

Example: -150deg is the same as 30deg

li {
  display: inline-block;
  width: calc((20px / 0.5) * var(--n,4)); /*cos(60deg) = 0.5*/
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  background-image: repeating-linear-gradient(-150deg, transparent, transparent 10px, black 10px, black 20px);
  background-size: 200% 100%;
  background-color: blue;
  animation: loading-slide 3s linear infinite;
}
.alt li {
  --n:6;
  background-image: repeating-linear-gradient(30deg, transparent, transparent 10px, black 10px, black 20px);
}

@keyframes loading-slide {
  from {
    background-position: left;
  }
  to {
    background-position: right;
  }
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

basically we add/remove 180deg until we get inside [-90deg,90deg] in order to be able to apply the formula.

Check this answer for more details about how background-size/background-position works: https://stackoverflow.com/a/51734530/8620333

Another Approach

Here is a complete different idea where you can rely on skew transformation and pseudo element. The trick here is that you don't have to define the width based on the stripes but the stripes will follow the width you will define so it will be easier to handle.

li {
  display: inline-block;
  width: calc( 20px * 3); /* it's only 20px * N */
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  position:relative;
  z-index:0;
  overflow:hidden
}
li::before {
  content:"";
  position:absolute;
  top:0;
  bottom:0;
  left:0;
  width:400%;
  /*we keep 0deg in the gradient*/
  background-image: repeating-linear-gradient(90deg, transparent, transparent 10px, black 10px, black 20px);
  transform:skewX(30deg);
  transform-origin:bottom left;
  animation: loading-slide 4s linear infinite;
}

@keyframes loading-slide {
  to {
    transform: translateX(-50%) skewX(30deg);
  }
}

.alt li {
  width: calc( 20px * 6);
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul class="alt">
  <li>test</li><li>test</li>
</ul>

As you can see, we keep a vertical gradient, we define the width of the element based on the width of the gradient. We make the pseudo element big enough and we apply a translation on it. The only thing you need to adjust is the skew transformation to control the degree.

With this approach you will also have better performance since you will animate transform instead of background-size.

More examples:

li {
  display: inline-block;
  width: calc( 20px * var(--n,3)); /* it's only 20px * N */
  height: 50px;
  margin-bottom:10px;
  background-color: blue;
  position:relative;
  z-index:0;
  overflow:hidden
}
li::before {
  content:"";
  position:absolute;
  top:0;
  bottom:0;
  left:-400%;
  right:-800%;
  /*we keep 0deg in the gradient*/
  background-image: repeating-linear-gradient(90deg, transparent, transparent 10px, black 10px, black 20px);
  transform:skewX(var(--d,30deg));
  animation: loading-slide 12s linear infinite;
}

@keyframes loading-slide {
  to {
    transform: translateX(-50%) skewX(var(--d,30deg));
  }
}

<ul>
  <li>test</li><li>test</li>
</ul>

<ul style="--n:6;--d:45deg">
  <li>test</li><li>test</li>
</ul>
<ul style="--n:8;--d:-70deg">
  <li>test</li><li>test</li>
</ul>
<ul style="--n:8;--d:30deg">
  <li>test</li><li>test</li>
</ul>

来源：https://stackoverflow.com/questions/54909275/animated-css-background-pattern-sliding-infinitely

标签

css

css-animations

linear-gradients

repeating-linear-gradient