在jQuery中获取所有没有子节点的元素 [英] Get all elements without child node in jQuery
问题描述
我需要选择没有子节点的元素(包括文本，因为在<p>中，文本是子节点).
I need to select elements without child node (including text since in <p> text is a child node).
我使用了empty，但它也将空间视为子节点.
I used empty, but it also consider space as child node.
示例:
标记:
<span> </span>
<span></span>
脚本:
$("span:empty").html("this was empty!");
不幸的是，由于第一个元素具有空格并且被视为子节点，因此只能选择和更改第二个元素.
Unfortunately, only the second element were selected and changed since the first element has space and it was considered child node.
如何选择没有子节点的元素?我想将空间视为空无一物.最好是，我希望代码不要使用循环来选择它们，可能还有其他方法.
How do I select elements without child node? I want to consider a space as nothing. Preferably, I want the code not to use loop to select them, there might be other ways.
推荐答案
如何
$("span:not(:has(*))")
选择所有没有子项的跨度.
Selects all spans that have no children.
说明
:has() 选择器选择的元素至少包含一个与指定的选择器."通配符*表示所有元素.
如果
<p>存在，则表达式$('div:has(p)')与<div>匹配 它的后代中的任何地方，不仅是直子.
The expression
$('div:has(p)')matches a<div>if a<p>exists anywhere among its descendants, not just as a direct child.
:not() 选择器选择与给定选择器不匹配的所有元素. "
The :not() selector "selects all elements that do not match the given selector."
在这种情况下，:has()选择所有内容，然后使用:not()查找与所有内容"都不匹配的元素……换句话说，没有任何内容.
In this case, :has() selects everything and then we use :not() to find the elements that don't match "everything"... in other words, nothing.
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