CSS选择器的第n个范围？ [英] CSS Selector for nth range?

问题描述

如何调整以下CSS选择器：

  .myTableRow td：nth-​​child（？）{
 background-color：#FFFFCC; 
} 
  

因此它适用于td列2-4？

 < table> 
< tr class =myTableRow> 
< td>列1< / td> 
< td>栏2< / td> 
< td> column 3< / td> 
< td> column 4< / td> 
< td>栏5< / td> 
< / tr> 
< / table> 
  

解决方案

单独：nth-​​child（） - 你需要链接至少一个其他这样的伪类。例如，：nth-​​child（）和：nth-​​last-child（） code> n + 2 位意味着分别从第二个孩子开始向前和向后计数）：

  .myTableRow td：nth-​​child（n + 2）：nth-​​last-child（n + 2）{
 background-color：#FFFFCC; 
} 
  

或者，不使用公式， c>：first-child 和：last-child ：

< .myTableRow td：not（：first-child）：not（：last-child）{
background-color：#FFFFCC;
}


How can I adapt the CSS selector below:

.myTableRow td:nth-child(?){
background-color: #FFFFCC;
}

so it applies to td columns 2-4?

<table>
<tr class="myTableRow">
<td>column 1</td>
<td>column 2</td>
<td>column 3</td>
<td>column 4</td>
<td>column 5</td>
</tr>
</table>

解决方案

You won't be able to do this with a single :nth-child() — you'll need to chain at least one other such pseudo-class. For example, a combination of :nth-child() and :nth-last-child() (the n+2 bit means start counting forward and backward respectively from the 2nd child):

.myTableRow td:nth-child(n+2):nth-last-child(n+2){
background-color: #FFFFCC;
}

Alternatively, instead of making use of a formula, simply exclude :first-child and :last-child:

.myTableRow td:not(:first-child):not(:last-child){
background-color: #FFFFCC;
}

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