组合：nth-​​of-type（）和：not [英] Combining :nth-of-type() and :not

问题描述

我有一个奇怪的情况试图结合两个伪类：：not 和：nth-​​of-type（）

I have a strange case of trying to combine two pseudoclasses: :not and :nth-of-type() to get rainbow striping on non-hidden items

有一个：not（[selector]）：nth-​​of-type ]），我假设css会先应用：not 项目，然后再应用：nth-​​of-type <

Having a :not([selector]):nth-of-type([rule]), I assumed css would filter the :not items first before applying the :nth-of-type rule, but having them in any order gives me the same result.

这里有一个jsfiddle演示了这个： http://jsfiddle.net/j7hjU/

Here's a jsfiddle that demonstrates this: http://jsfiddle.net/j7hjU/

我可能在做这些？

推荐答案

我假设css在应用之前先过滤：类型

I assumed css wouuld filter the :not items first before applying the :nth-of-type

不适用。 CSS是完全声明性的;每个选择器是一个独立于任何选择器部分的真或假的简单条件。这不是一个程序语言，你采取集和处理它，缩小每一步。具有过程规则的选择器语言将免于许多种优化，并且会更慢。

Nope. CSS is fully declarative; every selector is a simple condition that is true or false independently of any selector part. It's not a procedural language where you take a set and process it, narrowing it down with each step. A selector language with procedural rules would be immune to many kinds of optimisation and would be slower.

所以 nth-of-type 只是关于一个元素的父级中的位置，而不是在到目前为止的结果列表中的位置，因为CSS选择器没有这样的概念。选择器引擎可以在使用而不是缩小范围之前查找 nth-of-type 的测试不会相互干扰。

So nth-of-type is only about position within an element's parent, and not position in a 'results list so far' because CSS selectors have no such concept. A selector engine could look up the test for nth-of-type before narrowing it with not, as the rules do not interfere with each other.

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