[英]How to select an sub-element (se) of another element (e), but do not include one of sub-elements of selected sub-element (sse)?
Example: 例:
.element
|
|-.subelement
|
|-.subsubelement1
|-.subsubelement2
|-.subsubelement3
|-.non-desired-subsubelement
|-.subsubelement4
So how not to list all desired subsubelements (eg is unknown), but still receive an aggregated element .subelement
of its subelements .subsubelementX
without .non-desired-subsubelement
: 那么怎么不列出所有需要的subsubelements(如不明),但仍收到汇总元素
.subelement
及其子元素的.subsubelementX
没有.non-desired-subsubelement
:
NOT an array : [subsubelement1, subsubelement2, subsubelement3, subsubelement4]
不是数组 :
[subsubelement1, subsubelement2, subsubelement3, subsubelement4]
BUT an aggregate element : subelement
, without non-desired-subsubelement
as its descendant: BUT的集合体元件 :
subelement
,没有non-desired-subsubelement
作为其后代:
.subelement
|
|-.subsubelement1
|-.subsubelement2
|-.subsubelement3
|-.subsubelement4
I thought about this: 我想到了这个:
$('.element .subelement:not([class^="non-desired-subsubelement"])') // selects subelements which has no class non-desired-subsubelement
would be nice to have something like: 拥有这样的东西会很好:
$('.element .subelement:not(>>this<<:has([class^="non-desired-subelement"]))')
ie uses negative "has" selector; 即使用否定的“有”选择器; or something like:
或类似的东西:
$('.element .subelement:has-not([class^="non-desired-subelement"])') // much nicer
Or is it easier to get the .subelement
and than remove the .non-desired-subsubelement
, so to do it in two steps? 还是获得
.subelement
并比删除.non-desired-subsubelement
更容易 ,所以分两步进行?
var subelement = $('.element .subelement');
subelement.remove('.non-desired-subsubelement');
May be there is still a complex selector? 可能还有一个复杂的选择器?
您可以尝试以下方法:
$('.element .subelement [class^="sub"])
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