html - jQuery获取多个属性值

Question

我有以下 HTML 代码

<img src="" data-src="https://example.com/image1.jpg" class="img">
<img src="" data-src="https://example.com/image2.jpg" class="img">
<img src="" data-src="https://example.com/image3.jpg" class="img">
<img src="" data-src="https://example.com/image4.jpg" class="img">
<img src="" data-src="https://example.com/image5.jpg" class="img">
<img src="" data-src="https://example.com/image6.jpg" class="img">
<img src="" data-src="https://example.com/image7.jpg" class="img">

使用 JQuery，我想获取 data-src 值并将其放在 src

像这样的东西

$("*.img").attr("src", $(this).attr("data-src"));

感谢您的帮助！

标签： htmljqueryimageattributessrc

Answer 1

您可以使用.each来实现这一点。

  $("img").each((i, e) => {
    $(e).attr("src", $(e).attr("data-src"))
  });

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<img src="" data-src="https://example.com/image1.jpg" class="img">
<img src="" data-src="https://example.com/image2.jpg" class="img">
<img src="" data-src="https://example.com/image3.jpg" class="img">
<img src="" data-src="https://example.com/image4.jpg" class="img">
<img src="" data-src="https://example.com/image5.jpg" class="img">
<img src="" data-src="https://example.com/image6.jpg" class="img">
<img src="" data-src="https://example.com/image7.jpg" class="img">