jQuery - 重复ID的选择器 [英] jQuery - Selector for duplicate ID's
问题描述
我有一个页面,其中包含表单元素的重复ID。问题是我根据切换单独显示元素。因此,两个ID都不会同时显示。
I have a page with duplicate ID's for a form element. The catch is I the elements show up separately based on a toggle. So both ID's never show up simultaneously.
但是当我对该元素进行表单验证时,它总是选择代码中最后显示的元素(即使它隐藏了)。
However when I do form validation on that element, it always selects the element displayed last in the code (even if its hidden).
是否有选择器可以选择可见的重复ID?
Is there a selector to select the visible duplicate ID?
我尝试了以下但是没有可用:
I've tried the following but to no avail:
$('#my_element:visible').val();
推荐答案
由于关于此前提的无数其他问题将告诉我们在这种情况下,你不能使用ID选择器#
;你必须使用 $('div [id = foo]')
之类的东西来找到它。
As the myriad of other questions about this premise will tell you, you cannot use the ID selector #
in this case; you have to use something like $('div[id=foo]')
to find it.
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