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Javascript 使用ajax成功发送post值时如何显示此div?_Javascript_Jquery_Html_Css_Ajax - Fatal编程技术网

Javascript 使用ajax成功发送post值时如何显示此div?

Javascript 使用ajax成功发送post值时如何显示此div?,javascript,jquery,html,css,ajax,Javascript,Jquery,Html,Css,Ajax,在一篇AJAX文章成功之后,如何显示以下div 我想展示这个 <div class="love" id="love_0" style="border-radius: 3px; padding: 8px; border: 1px solid #ccc; right: 13px; background: #fff; top: 13px;"> <a class="like" style="cursor: pointer;" id="876876"> &l

在一篇AJAX文章成功之后,如何显示以下div

我想展示这个

<div class="love" id="love_0" style="border-radius: 3px; padding: 8px; border: 1px solid #ccc; right: 13px; background: #fff; top: 13px;">
    <a class="like" style="cursor: pointer;" id="876876">
        <img src="dislove.png" style="border: none;" />
    </a>            
</div>

在ajaxsuccess函数中,您只需要将想要的div附加到它的容器中,以用于exbody

success: function(){            
   $('#'+uid).html('<img src="dislove.png" style=" border: none;">').addClass('unlike').removeClass('like');
   $('#you'+uid).text('');
   $('body').append('<div class="love" id="love_0" style=" border-radius: 3px; padding: 8px;  border: 1px solid #ccc; right: 13px; background: #fff; top: 13px; "><a class="like" style=" cursor: pointer; " id="876876"><img src="dislove.png" style=" border: none;"></a>            </div>')
}  

我不确定您是否在键入时出错,但var postData='uid='+uid';有一个额外的'将打破您的代码的其余部分。–詹姆斯海-谢谢,这只是示例代码^^
success: function(){            
   $('#'+uid).html('<img src="dislove.png" style=" border: none;">').addClass('unlike').removeClass('like');
   $('#you'+uid).text('');
   $('body').append('<div class="love" id="love_0" style=" border-radius: 3px; padding: 8px;  border: 1px solid #ccc; right: 13px; background: #fff; top: 13px; "><a class="like" style=" cursor: pointer; " id="876876"><img src="dislove.png" style=" border: none;"></a>            </div>')
}