Question

我有这个html结构：

<select class="required form-control parameter-criteria" selectpicker="" data-id="2" tabindex="-98">
        <option value="1" data-content="<span class=&quot;badge-min filter d-inline-flex&quot; style=&quot;background-color:#ff0000;&quot;> </span>CRITERIO 1 [0 - 50] - CRITERIO 2" id-criteria="2">CRITERIO 1
        </option>
</select>
<select class="required form-control parameter-criteria" selectpicker="" data-id="2" tabindex="-98">
        <option value="1" data-content="<span class=&quot;badge-min filter d-inline-flex&quot; style=&quot;background-color:#ff0000;&quot;> </span>CRITERIO 1 [0 - 50] - CRITERIO 2" id-criteria="2">CRITERIO 1
        </option>
</select>

我正在尝试选择一个特定的选择，所以我这样做了：

var select = $('.parameter-criteria[data-id=2]').eq(2);

这应该只返回最后选择的.eq(2)，但是如果我打印select会得到两个，为什么？

Answer 1

这是因为eq()索引是基于0的……就像第一个元素为0，第二个元素为1，依此类推。您只有两个匹配的元素，但是eq(2)在列表中寻找不存在的第三个孩子并返回全部。

var select = $('.parameter-criteria[data-id=2]').eq(0);
console.log(select.text().trim());
var select2 = $('.parameter-criteria[data-id=2]').eq(1);
console.log(select2.text().trim());

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<select class="required form-control parameter-criteria" selectpicker="" data-id="2" tabindex="-98">
        <option value="1" data-content="<span class=&quot;badge-min filter d-inline-flex&quot; style=&quot;background-color:#ff0000;&quot;> </span>CRITERIO 1 [0 - 50] - CRITERIO 2" id-criteria="2">CRITERIO 1
        </option>
</select>
<select class="required form-control parameter-criteria" selectpicker="" data-id="2" tabindex="-98">
        <option value="1" data-content="<span class=&quot;badge-min filter d-inline-flex&quot; style=&quot;background-color:#ff0000;&quot;> </span>CRITERIO 1 [0 - 50] - CRITERIO 2" id-criteria="2">CRITERIO 2
        </option>
</select>

.eq返回全部选择

1 个答案: