Question

我试图将所有来自特定元素的Div包裹起来。我怎么能用jQuery做到这一点？

当前代码

<div class="a"></div>
<div class="b"></div>
<div class="b"></div>

<div class="a"></div>
<div class="b"></div>
<div class="b"></div>

<div class="a"></div>
<div class="b"></div>
<div class="b"></div>

所需输出

<div class="a"></div>
<div class="wrapper">
    <div class="b"></div>
    <div class="b"></div>
</div>

<div class="a"></div>
<div class="wrapper">
    <div class="b"></div>
    <div class="b"></div>
</div>

<div class="a"></div>
<div class="wrapper">
    <div class="b"></div>
    <div class="b"></div>
</div>

Answer 1

您可以为每个SELECT pd.name AS 'Product Name', p.model AS UPC, p.quantity AS 'Quantity', p.price AS 'Regular Price', ps.price AS 'Special Price', p.cost AS 'COST', p.status AS 'Status' FROM oc_product p INNER JOIN oc_product_description pd ON pd.product_id = p.product_id INNER JOIN oc_product_special ps ON ps.product_id = p.product_id INNER JOIN oc_manufacturer m ON p.manufacturer_id = m.manufacturer_id INNER JOIN oc_product_to_category p2c ON p2c.product_id = p.product_id INNER JOIN oc_category c ON c.category_id = p2c.category_id INNER JOIN oc_category_description cd ON c.category_id = cd.category_id WHERE c.category_id = 37 OR c.parent_id = 37 GROUP BY pd.name ORDER BY m.name ASC创建wrap元素，将所有元素添加到下一个.a元素，然后添加到DOM。

.a

$('.a').each(function() {
  var wrap = $('<div class="wrapper"></div>');
  $(this).nextUntil('.a').appendTo(wrap);
  $(this).after(wrap);
})

.wrapper {
  border: 1px solid black;
}

Answer 2

如果总有2个.b，您可以使用以下代码：

$('.a').each(function() {
    $(this).next('.b').next('.b').andSelf().wrapAll('<div class="wrapper"/>');
});

.wrapper {
  border: 1px solid black;
  }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="a">A</div>
<div class="b">B</div>
<div class="b">B</div>

<div class="a">A</div>
<div class="b">B</div>
<div class="b">B</div>

<div class="a">A</div>
<div class="b">B</div>
<div class="b">B</div>

使用jQuery在某个元素之后使用div包装所有元素

2 个答案: