Question

我正在尝试使用jQuery和AJAX来表示喜欢，当我点击表情符号时，该值在数据库中就会增加。但是我必须刷新页面才能看到更新的值。

最初，我以为id是有问题的，我了解到HTML中的id属性应该以[a-zA-Z]开头。我修好了现在无法解决计数器动态更新的问题。

下面是要显示类似计数器的div。

<div class="article-left-box">
  <div class="article-left-box-inner">
    <div class="article-share">
      <a href="#" class="facebook"></a>
      <a href="#" class="twitter"></a>
    </div>
  <span class="add-to-favorite" data-zebra-tooltip title="Ad to favorite">
    <i class="material-icons">&#xE866;</i>
  </span>
 <ul class="article-emoticons">
   <li><a class="popular happy"></a><span><?php echo $row["total_reactions"]; ?></span>
     <ul>
       <li><a id=<?php echo $row['news_id']?> class="love" style='cursor:pointer;'></a><span><?php echo $row["love_reactions"]; ?></span></li>
       <li><a  class="shocked" style='cursor:pointer;'></a><span><?php echo $row["shocked_reactions"]; ?></span></li>
     </ul>
   </li>                                        
 </ul>

</div>
</div>

下面是jQuery

$(document).ready(function(){
  $(".love").click(function(){
    var idofpost=$(this).attr('id');
    //alert(idofpost);
    $.ajax({
    type:'POST',
    url:'reactions.php',
    data:'id='+idofpost,
    success:function(data)
    {
      $(this).children().first().html(data);
    }
   });
 });
});

下面是Reactions.php

<?php
  include 'dbconnection.php';
  $id = $_POST['id'];
  $sql = "UPDATE news_mst SET love_reactions = love_reactions + 1 WHERE news_mst.news_id = '$id'";
  if (!$result = $connection->query($sql)) {
    die ('There was an error running query[' . $connection->error . ']');
  } 

  $sql = "SELECT love_reactions from news_mst WHERE news_mst.news_id = '$id'";
  if (!$result = $connection->query($sql)) {
    die ('There was an error running query[' . $connection->error . ']');
  }
  //$result = mysqli_query($link,$sql);
  $row = mysqli_fetch_array($result);
  echo "$row[0]";
?>

即使AJAX调用成功，为什么数据不能实时更新？

0 个答案: