weixin_33708432
2020-04-20 17:28
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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>

图片转代码服务由CSDN问答提供 功能建议

                    

我假设我不太了解jquery和ajax,所以我尝试尝试创建“ like”按钮。 我现在使用codeigniter 4,目前我已经在控制器中创建了一个简单的函数,传递了帖子ID并输入了查询(这很简单,因为我现在需要理解并使其起作用)。

但是,不幸的是,我遇到了各种各样的问题,我相信它不会通过放置在中的data-id传递ajax代码中的数据。

我在做什么错? 或更好,我应该添加什么?

  $('。like')。on(“ click”,function(){
     var post_id = $('a')。attr('data-id');
     $ .ajax({
         输入:“ POST”,
         网址:“ <?= site_url('auth / sendVote');  ?>',
         数据:“ post_id =“ + post_id,
         快取:false,
         成功:功能(响应){
             $()。html(response +“”);
         }
     });
 });