Php jquery append未正确附加样式化图标

因此，我从一个PHP站点获得了一段html代码，返回以下内容：

<span class="inline-icon link-icon with-fav" style="background: url("/system/files/imagecache/s_favicon/attachments/links/favicons-201203/1d5920f4b44b27a802bd77c4f0536f5a_48.png?1332860368") no-repeat scroll 0% 0% transparent;"></span>

但是图标没有正确显示，当我使用firebug检查元素时，它显示如下：

<span class="inline-icon link-icon with-fav" transparent;"="" 0%="" scroll="" no-repeat="" 1d5920f4b44b27a802bd77c4f0536f5a_48.png?1332860368")="" favicons-201203="" links="" attachments="" s_favicon="" imagecache="" files="" system="" style="background: url("></span>

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一旦应用了语法着色，就很容易发现可能出错的地方——您需要在内联样式中使用单引号，而不是嵌套的双引号

<span class="inline-icon link-icon with-fav" style="background: url('/system/files/imagecache/s_favicon/attachments/links/favicons-201203/1d5920f4b44b27a802bd77c4f0536f5a_48.png?1332860368') no-repeat scroll 0% 0% transparent;"></span>

更好的方法是使用外部类而不是内联样式

<span class="inline-icon link-icon with-fav" style="background: url('/system/files/imagecache/s_favicon/attachments/links/favicons-201203/1d5920f4b44b27a802bd77c4f0536f5a_48.png?1332860368') no-repeat scroll 0% 0% transparent;"></span>