Question

我有一个我想替换的图片但是......

<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>

我已经使用此代码尝试替换整个img标记，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">" );

有人可以告诉我这段代码有什么问题吗？

Answer 1

请勿使用replacewith替换图像。只需更改src：

即可

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

Answer 2

您的代码语法无效。你应该转义字符串中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');

Answer 3

你在双引号内使用双引号，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 4

好吧，您可以尝试转义引号，例如"<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\">" );

或类似'<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">');

如何使用.replaceWith替换图像源？

4 个答案: