使用jQuery [duplicate]获取最后一个可见元素

This question already has an answer here:

这个问题在这里已有答案:

jquery selector unable to find visible last-child 3 answers

jquery选择器无法找到可见的last-child 3答案

<table>
    <tr class="here" id="t1" number="1" ><td>1</td></tr>
    <tr class="here" id="t2" number="2" ><td>2</td></tr>
    <tr class="here" id="t3" number="3" style="display:none"><td>3</td></tr>
    <tr class="here" id="t4" number="4" style="display:none"><td>4</td></tr>
</table>

<span id="check">check</span>

$('#check').click(function(){        
    check = ???;
   alert(check);
})

DEMO: http://jsfiddle.net/vUukc/1/

How can I get the attribute number from last visible tr in this example? This is an example - all <tr> could be visible.

如何从此示例中的最后一个可见tr获取属性编号?这是一个例子 - 所有都可见。

3 个解决方案

#1

There are different jQuery Selectors for those purposes.

出于这些目的,有不同的jQuery选择器。

For example:

$('table tr:visible:last').attr('number');

$('table tr:visible').last().attr('number');

and so on.

等等。

Full code

$('#check').click(function() {
    check = $('table tr:visible:last').attr('number');
    alert(check);
});

DEMO

#2

$("table tr:visible:last").attr("number");

See the jQuery Documentation on Selectors for more info on the :visible and :last selectors.

有关:visible和:last选择器的更多信息,请参阅选择器上的jQuery文档。

#3

Try this

Live demo

   $('#check').click(function(){

    alert($('table').find('tr:visible:last').attr("number"));

   });

#1