我有以下代码:

for i in (1..5)
 div:nth-child({i})::after
  content \'i*i\'

for i in (6..10)
 div:nth-child({i})::after
  ij = (i - 5)
  content \'(ij*ij)\'

for i in (1..10)
  div:nth-child({i})
  ij = (i - 5)
  if i >= 6
   height ij * ij px
  else
   height i * i px


输出以下内容:



但是我需要在25时反转,所以它会:

1 4 9 16 25 25 16 16 9 4 1

不知道该怎么做?

最佳答案

将计算ij的方式从ij =(i-5)更改为ij = 11-i

for i in (1..5)
 div:nth-child({i})::after
  content \'i*i\'

for i in (6..10)
 div:nth-child({i})::after
  ij = 11-i
  content \'(ij*ij)\'

for i in (1..10)
  div:nth-child({i})
  ij = 11-i
  if i >= 6
   height ij * ij px
  else
   height i * i px

关于css - 如何在中途反转循环?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/24996135/

10-12 13:13