Question

有没有办法在这种代码（sass）中制作css的抽象规则：

#cottage-image-gallery input:nth-of-type(1):checked ~ label:nth-of-type(1) img,
#cottage-image-gallery input:nth-of-type(2):checked ~ label:nth-of-type(2) img,
#cottage-image-gallery input:nth-of-type(3):checked ~ label:nth-of-type(3) img,
#cottage-image-gallery input:nth-of-type(4):checked ~ label:nth-of-type(4) img,
#cottage-image-gallery input:nth-of-type(5):checked ~ label:nth-of-type(5) img,
#cottage-image-gallery input:nth-of-type(6):checked ~ label:nth-of-type(6) img,
#cottage-image-gallery input:nth-of-type(7):checked ~ label:nth-of-type(7) img,
#cottage-image-gallery input:nth-of-type(8):checked ~ label:nth-of-type(8) img
  position: fixed

可以变成这样的东西：

#cottage-image-gallery input:nth-of-type(n):checked ~ label:nth-of-type(n) img
  position: fixed

当n等于1时，第二个变量相同;当n等于2时，第二个变量变为2;等等...

我没有使用相邻选择器＆＃34; +＆＃34;是因为我需要在同一个父母下面进行输入，但所有输入都在彼此附近。

最好的问候

示例：https://codepen.io/s3m3nT3s/pen/rYBxLq

Answer 1

在Sass中，您可以使用@for directive来执行此操作。

@for $i from 1 through 8
  #cottage-image-gallery input:nth-of-type(#{$i}):checked ~ label:nth-of-type(#{$i}) img
    position: fixed

输出：

#cottage-image-gallery input:nth-of-type(1):checked ~ label:nth-of-type(1) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(2):checked ~ label:nth-of-type(2) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(3):checked ~ label:nth-of-type(3) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(4):checked ~ label:nth-of-type(4) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(5):checked ~ label:nth-of-type(5) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(6):checked ~ label:nth-of-type(6) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(7):checked ~ label:nth-of-type(7) img {
  position: fixed;
}
#cottage-image-gallery input:nth-of-type(8):checked ~ label:nth-of-type(8) img {
  position: fixed;
}

但是，给定这样的HTML：

<input id="slide1" type="radio" name="cottage-image" data="1">
<input id="slide2" type="radio" name="cottage-image">
<input id="slide3" type="radio" name="cottage-image">
<input id="slide4" type="radio" name="cottage-image">
<input id="slide5" type="radio" name="cottage-image">
<input id="slide6" type="radio" name="cottage-image">
<input id="slide7" type="radio" name="cottage-image">
<input id="slide8" type="radio" name="cottage-image">
<input id="slide0" type="radio" name="cottage-image" checked>
<label for="slide1"><img src="http://calhaugrande.com/img/sol/1.jpg"></label>
<label for="slide2"><img src="http://calhaugrande.com/img/sol/2.jpg"></label>
<label for="slide3"><img src="http://calhaugrande.com/img/sol/3.jpg"></label>
<label for="slide4"><img src="http://calhaugrande.com/img/sol/4.jpg"></label>
<label for="slide5"><img src="http://calhaugrande.com/img/sol/5.jpg"></label>
<label for="slide6"><img src="http://calhaugrande.com/img/sol/6.jpg"></label>
<label for="slide7"><img src="http://calhaugrande.com/img/sol/7.jpg"></label>
<label for="slide8"><img src="http://calhaugrande.com/img/sol/8.jpg"></label>
<label for="slide0"></label>

在纯CSS中根本没办法将input[id="slide1"]与label[for="slide2"]，input[id="slide2"]与label[for="slide2"]匹配，依此类推，而不是像你一样重复自己很多次是:nth-child()。

CSS的方法是这样的：

#cottage-image-gallery input:nth-of-type([id]):checked ~ label:nth-of-type([for]) img

但是你不能在:nth-child()中使用属性选择器。也许在将来！

摘要CSS

1 个答案: