如何使用 .replaceWith 替换图像的源

Question

我有一个图像我想替换，但是...

<div class="news-thumb-wrapper">
     <img src="/content/oldimage.jpg" class="attachment-post-thumbnail" height="150" width="600">
     <div class="gallery-arrows"></div><h8><a href="/">Some text</a></h8></div>

我使用此代码尝试替换整个 img 标签，但它似乎不起作用。

$( "img.attachment-post-thumbnail" ).replaceWith( "<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">" );

谁能告诉我代码有什么问题？

Answer 1

不要使用 replacewith 来替换图像。只需更改src：

$("selector for img").attr("src", "/content/newimage.jpg");

Answer 2

你在Double

Cote中使用Double Cote，你应该做的是：

$( "img.attachment-post-thumbnail" ).replaceWith('<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">')

Answer 3

好吧，您可以尝试像这样转义引号"<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">"

$( "img.attachment-post-thumbnail" ).replaceWith( "<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">" );

或者像那样'<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">'

$( "img.attachment-post-thumbnail" ).replaceWith('<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">');

Answer 4

您的代码语法无效。您应该转义字符串内的双引号或使用单引号：

$( "img.attachment-post-thumbnail" ).replaceWith( '<img src="/content/newimage.png" class="attachment-post-thumbnail" height="41" width="600">' );

现在，交换图像的正确方法是简单地设置其src属性：

$( "img.attachment-post-thumbnail" ).prop('src', '/content/newimage.png');