load() PHP 在元素之后

Question

因为某些原因我需要通过jquery加载一个PHP文件（在最后一点它将负责调用wordpress动态侧边栏但我无法在里面执行php jquery)。我现在拥有的是：

$('#pa_kategoria-cenowa').after('<div class="product-variation-desc"><div id="5" class="box">CENA 5 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div><div id="10" class="box">CENA 10 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div><div id="15" class="box">CENA 15 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div></div>');

我的php文件被调用：load-widgets.php

这就是加载函数应该看起来像这样的原因：

.load('load-widgets.php');

PHP 文件的内容如下：

<div class="product-variation-desc"><div id="5" class="box">CENA 5 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div><div id="10" class="box">CENA 10 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div><div id="15" class="box">CENA 15 - Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aenean semper egestas sodales. In commodo massa eget lectus eleifend, nec dictum sapien laoreet. </div></div>

但我不知道如何将它们组合在一起。我试过：

$('#pa_kategoria-cenowa').after().load('load-widgets.php');

但它不起作用。如何实现？

Answer 1

您现有的代码应该可以工作。检查控制台中的错误。如果仍未解决，请尝试以下 AJAX 代码段。

$.get("load-widgets.php", function(data){ // Loads content into the 'data' variable.
   $("#pa_kategoria-cenowa").after(data); 
});