我试图弄清楚如何遍历所有父类(.grid),如果它没有具有类(.image-container)的子div,然后在同一(.grid)部分中显示(.content-container) 。

HTML:

<style>
.grid .content-container {
    display:none;
}
</style>

<div class="grid">
    <div class="art">
        <div class="image-container">
            <img src="image url" />
        </div>
        </div>
        <div class="title">Title Text</div>
        <div class="content-container">Some Content Goes here</div>
</div>
<div class="grid">
    <div class="art"></div>
        <div class="title">Title Text</div>
        <div class="content-container">Some Content Goes here</div>
</div>
<div class="grid">
    <div class="art">
        <div class="image-container">
            <img src="image url" />
        </div>
        </div>
        <div class="title">Title Text</div>
        <div class="content-container">Some Content Goes here</div>
</div>
<div class="grid">
    <div class="art">
        <div class="image-container">
            <img src="image url" />
        </div>
        </div>
        <div class="title">Title Text</div>
        <div class="content-container">Some Content Goes here</div>
</div>
<div class="grid">
    <div class="art"></div>
        <div class="title">Title Text</div>
        <div class="content-container">Some Content Goes here</div>
</div>

最佳答案

这样的事情应该起作用:

$('.grid').each(function() {
    if($(this).find('.image-container').length == 0) {
        // no children
        $(this).find('.content-container').show();
    }
});

关于jquery - 遍历父元素,如果子元素不存在,则显示文本,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/25071441/

10-11 12:52
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