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by Sinescape
1126 days ago
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This combination feels unnecessarily complicated to me, at least for most of the use cases shown. Suppose we want to style a list differently if it has five or more items. Since :has can detect any child element, the existence of a fifth item is a sufficient indicator that there are at least five items. So instead of ul:has(li:nth-last-child(n+5)), we could just use ul:has(li:nth-child(5)) which is quite a bit clearer in my opinion. |
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