Question

我有以下html结构（这只是一个概述，所有div都有内容并且已正确关闭）。每个div.views-row包含一个views-field-title，views-field-body和views-field-afbeeldingen

<div class="view view-toepassingen view-id-toepassingen view-display-id-page view-dom-id-5a6ac8323a7566e5f11218e7b6c49c5c">
    <div class="view-content">
    <div class="views-row views-row-1 views-row-odd views-row-first">
        <div class="views-field views-field-title">
        <div class="views-field views-field-body">
        <div class="views-field views-field-field-afbeeldingen">
    </div>
    <div class="views-row views-row-2 views-row-even">
    <div class="views-row views-row-3 views-row-odd">
    <div class="views-row views-row-4 views-row-even">
    <div class="views-row views-row-5 views-row-odd">
    <div class="views-row views-row-6 views-row-even views-row-last">
</div>

我想要实现的是将div重新排列为:(有效地移动'afbeeldingen'div内的标题和正文div，以及每个视图行的内容）

<div class="view view-toepassingen view-id-toepassingen view-display-id-page view-dom-id-5a6ac8323a7566e5f11218e7b6c49c5c">
    <div class="view-content">
    <div class="views-row views-row-1 views-row-odd views-row-first">
        <div class="views-field views-field-field-afbeeldingen">
            <div class="views-field views-field-title">
            <div class="views-field views-field-body">
        </div>
    </div>
    <div class="views-row views-row-2 views-row-even">
    <div class="views-row views-row-3 views-row-odd">
    <div class="views-row views-row-4 views-row-even">
    <div class="views-row views-row-5 views-row-odd">
    <div class="views-row views-row-6 views-row-even views-row-last">
</div>

这是我的jquery代码行为不端。我正试图遍历所有6个视图行div。

var i = 1;
$('.view-toepassingen > .view-content').children('div').each(function () {
    $('.view-toepassingen > .view-content > .views-field-field-afbeeldingen').prepend($('.view-toepassingen > .view-content > .views-row-'+i +' .views-field-title'));
    $('.view-toepassingen > .view-content > .views-field-field-afbeeldingen').prepend($('.view-toepassingen > .view-content > .views-row-'+i +' .views-field-body'));
    i++;
});

Answer 1

在.each回调中，this引用当前的div项。您可以使用$(this).find()搜索此节点的子节点：

$('.view-toepassingen > .view-content').children('div').each(function () {
    var $title = $(this).find('.views-field-title');
    var $body = $(this).find('.views-field-body');

    $(this).find('.views-field-field-afbeeldingen').append($title, $body);
});

.find()会调查此节点的所有儿童（在任何深度），.children()只会查看其直接子女等...

Answer 2

在each() div上执行afbeeldingen并搜索并移动每个div的兄弟姐妹可能更有意义：

$(".views-field-field-afbeeldingen").each(function() {
    $(this).siblings(".views-field-title").appendTo(this);
    $(this).siblings(".views-field-body").appendTo(this);
});

另请注意，使用适当的pseudo classes可以更轻松地实现您拥有的所有手动奇数，偶数，第一，最后一个课程。

请参阅this fiddle。

用Jquery重新排列div

2 个答案: