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/