javascript - 获取每行中也与复合选择器匹配的第 n 个子节点

Question

假设我有以下标记

<div><p></p><blockquote></blockquote><ul>...</ul></div>
<div><p></p><blockquote></blockquote><ul>...</ul></div>
 ...
<div><p></p><blockquote></blockquote><ul>...</ul><ul></ul></div>

有没有我可以使用的 jQuery 选择器/遍历技术来让第 n 个子节点匹配每个 div 中的复杂选择器，这是健壮的(从某种意义上说，我可以更改与选择器不匹配的元素而不会把事情搞砸)

以下是一些无法帮助说明问题的方法

 $("div").children("p, ul").filter(":eq(1)") // returns just the first <ul> in the first div
 $("div").children("p, ul:eq(1)") // returns all the p's and the first ul in the first div
 $("div").find(":not(blockquote):eq(1)") //returns the correct elements, but at the expense of having to reference the elements we're *not* after. Also not restricted to just children
 $("div").find(">:not(blockquote):eq(1)") ) // tackles the children problem, but ">" without a parent is not officially supported and jQuery want to deprecate (Can't find the reference but I raised it once in discussion on the jQuery bug boards and John Resig himself said this)

最后，对于某些上下文，我问这个是为了始终找到表的第 i 列，而不管第一列是否包含 th 或 td细胞

Answer 1

是的，jQuery 有一个nth-child 选择器

jQuery - :nth-child() Selector