weixin_33708432 2020-04-20 17:28 采纳率: 0%
浏览 35

Ajax投票系统

I assume that I don't know a lot of jquery and ajax, I'm venturing trying to be able to create a "like" button. I use codeigniter 4, for the moment I have created a simple function in the controller, passing the post ID and entering the query (it's all simple because I need to understand and make it work for the moment).

Unfortunately, however, I have various problems, I believe that it does not pass the data in the ajax code via the data-id placed in

What am I doing wrong? or better, what should I add?

$('.like').on("click",function(){
    var post_id = $('a').attr('data-id');
    $.ajax({
        type: "POST",
        url: '<?= site_url('auth/sendVote'); ?>',
        data: "post_id="+post_id,
        cache: false,
        success: function(response){
            $().html(response+"");
        }
    });
});



<a href="" class="like card-btn" data-id="<?= $item['id']; ?>">
<div class="total ml-auto">
     <?= $model->countVotes($item['id']); ?>
</div>
</a>
  • 写回答

0条回答 默认 最新

    报告相同问题?

    悬赏问题

    • ¥100 Jenkins自动化部署—悬赏100元
    • ¥15 关于#python#的问题:求帮写python代码
    • ¥20 MATLAB画图图形出现上下震荡的线条
    • ¥15 关于#windows#的问题:怎么用WIN 11系统的电脑 克隆WIN NT3.51-4.0系统的硬盘
    • ¥15 perl MISA分析p3_in脚本出错
    • ¥15 k8s部署jupyterlab,jupyterlab保存不了文件
    • ¥15 ubuntu虚拟机打包apk错误
    • ¥199 rust编程架构设计的方案 有偿
    • ¥15 回答4f系统的像差计算
    • ¥15 java如何提取出pdf里的文字?