在jQuery中有样式选择器吗? [英] Is there a style selector in jQuery?
问题描述
如果我想选择其中的所有图片,例如Home,可以这样做:
$ (img [alt ='Home'])
但是如何选择
编辑:没有这样的选择器,有没有任何插件,或任何计划,在下一个jQuery版本?
一个好主意,但你可以为它添加一个新的Sizzle选择器:
$。expr [':']。width = function(elem,pos,match){
return $(elem).width()== parseInt(match [3]);
}
您可以这样使用:
$('div:width(970)')
b $ b
这样会变得非常缓慢,因此你可以缩小要比较的元素数量,例如:
$('#navbar> div:width(970)')
仅选择作为导航条直接后代的div,其宽度也为970像素。
If I want to select every image which it's alt is Home for example, I can do something like this:
$("img[alt='Home']")
But how can I select every elements which their width
CSS property is 750px for example in a single selector?
EDIT: If there is no such selector, is there any plugin, or any plans to do it in the next jQuery versions?
Not necessarily a great idea, but you could add a new Sizzle selector for it :
$.expr[':'].width = function(elem, pos, match) {
return $(elem).width() == parseInt(match[3]);
}
which you could then use like so:
$('div:width(970)')
That's going to be horrifically slow, though, so you'd want to narrow down on the number of elements you're comparing with something like :
$('#navbar>div:width(970)')
to only select those divs that are direct descendants of the navbar, which also have a width of 970px.
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