我正在尝试创建一个动画重复模式(对角线条纹水平滑动),作为加载块的占位符(本例中为li
)。
如何使动画光滑/连续,给幻象以无限滑动的模式?
width
,使模式是连续的?(条纹不应显得破损/中断)。目标是有一个类,我可以添加到任何块,并在视觉上看起来像加载/处理。
注释:Note;纯。
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>
发布于 2019-02-27 07:56:27
正确的公式应该是(20px / cos(45deg)) * N
。然后,您可以将background-size
设置为200% 100%
(比元素大两倍),并将其从左向右动画:
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>
你可以考虑任何程度,并根据需要调整公式。(20px / cos(90deg - |Xdeg|)) * N
与X
在-90deg
和90deg
之间
使用-60deg
的示例
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>
使用30deg
的示例
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>
使用80deg
的示例
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>
您可以清楚地识别X=+/-90deg
(垂直条纹)和cos(0)=1
的简单情况,因此公式将是20px * N
。同样,当X=0
(水平条纹),我们将有cos(90deg) = 0
和任何宽度将工作,因为没有垂直模式(公式不再定义)
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>
[-90deg,90deg]
**?**以外的值是什么?
上面的范围已经涵盖了180deg
,而且由于我们正在处理一些系统,所以所有的值都可以在该范围内表示。
示例:110deg
与-70deg
相同
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>
示例:-150deg
与30deg
相同
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>
基本上,我们添加/删除180deg
,直到我们进入[-90deg,90deg]
,以便能够应用公式。
有关background-size
/background-position
如何工作的更多详细信息,请查看此答案:Using percentage values with background-position on a linear-gradient
另一种方法
这里有一个完全不同的想法,您可以依赖于skew
转换和伪元素。这里的诀窍是,您不必根据条纹来定义宽度,但是条纹将遵循您定义的宽度,这样就更容易处理了。
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>
如您所见,我们保持垂直梯度,根据梯度的宽度定义元素的宽度。我们使伪元素足够大,并在其上应用一个转换。唯一需要调整的是控制度的倾斜转换。
使用这种方法,您也将有更好的性能,因为您将动画转换,而不是background-size
。
更多的例子:
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>
https://stackoverflow.com/questions/54909275
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