Question

我有一些DOM结构：

<div class="model-A">
    <div class="some1">
        <div class="sub1"></div>
        <div class="model-B">
            <div class="some2">
                <div class="sub2"></div>
            </div>
            <div class="model-C">
                <div class="some3">
                    <div class="sub3"></div>
                    <div class="model-B">
                        <div class="some4"></div>
                        <div class="sub4"></div>
                    </div>
                </div>
            </div>
            <div class="some2">
                <div class="sub2"></div>
            </div>
        </div>
    </div>
</div>

如何获得具有＆＃39;模型-B＆＃39;的div的所有元素？上课，但不是其他模特 - *＆＃39;孩子的？例如，结果必须是包含元素：

从第一个＆＃39;模型-B＆＃39;

<div class="some2">
   <div class="sub2"></div>
</div>
<div class="some2">
   <div class="sub2"></div>
</div>

从第二个＆＃39;模型-B＆＃39;

<div class="some4"></div>
<div class="sub4"></div>

Answer 1

您可以使用:has选择器执行此操作，如下所示：

$('.model-B>div').not(':has([class^=model])')

演示：

＆＃13;

$('.model-B>div').not(':has([class^=model])').each(function () {
    console.log(this.outerHTML);
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="model-A">
    <div class="some1">
        <div class="sub1"></div>
        <div class="model-B">
            <div class="some2">
                <div class="sub2"></div>
            </div>
            <div class="model-C">
                <div class="some3">
                    <div class="sub3"></div>
                    <div class="model-B">
                        <div class="some4"></div>
                        <div class="sub4"></div>
                    </div>
                </div>
            </div>
            <div class="some2">
                <div class="sub2"></div>
            </div>
        </div>
    </div>
</div>

＆＃13;

如何获得嵌套元素？

1 个答案: