Question

在下面的代码中，我正确地从数据库中获取了所有数据。但接下来我想通过使用动态生成的删除链接来删除特定数据..但是我无法获得特定的ID以便在delete.php页面上发送它。我总是得到未定义的ID，这是我在alert.please帮助我如何获取我的myId变量中的Id发送到下一页。

 <script>
    $(document).ready(function(){
       var myId=$(this).data('Id');
       alert(myId);
      $(".delete").click(function(){
          $.ajax({
             type:"Post",
             url:"delete.php",
             data:{
                 Id:myId,
              },
             sucess:function(data){

            }
            });
        });
    });
</script>
 <?php

  $localhost="localhost";
  $username="root";
  $password="dheeraj";
  $dbname="mydatabase";

  $con = mysqli_connect('localhost','root','dheeraj','mydatabase');
  if(!$con){
  echo mysqli_error($con); 
   }

    $action=$_POST['action'];

    if($action=='showComment'){
      $sql="select * from employee order by Id";
      $result=mysqli_query($con,$sql);
     if(mysqli_num_rows($result)>0){
        echo "<table border=1px solid red><tr><th>Id</th><th>FirstName</th><th>LastName</th><th>Email</th><th>Contact Number<th>Delete</th></th></tr>";
      while($row=mysqli_fetch_array($result)){

           echo '<tr><td>'.$row["Id"].'</td><td>'.$row["Fname"].'</td><td>'.$row["Lname"].'</td><td>'.$row["email"].'</td><td>'.$row["contact"].'</td><td><a class="delete" href="?id="'.$row["Id"].'>Delete</a></td></tr>';
    }
    echo "</table>";
 }
}
?>

Answer 1

您的删除链接元素中没有定义数据属性，请输入：

<a class="delete" href="?id="'.$row["Id"].' data-Id="'.$row["Id"].'">Delete</a>

你的js应该是：

$(document).ready(function(){
  // replace document below with top level parent that already existed on page
  $(document).on('click', '.delete', function(){
    var myId  =$(this).data('Id');
    alert(myId);
      $.ajax({
         type:"Post",
         url:"delete.php",
         data:{
             Id:myId,
         },
         sucess:function(data){

         }
        });
    });
});

Answer 2

尝试按如下方式重写PHP代码：

if(mysqli_num_rows($result)>0){
        echo "<table border=1px solid red><tr><th>Id</th><th>FirstName</th><th>LastName</th><th>Email</th><th>Contact Number<th>Delete</th></th></tr>";
      while($row=mysqli_fetch_array($result)){

           echo '<tr><td>'.$row["Id"].'</td><td>'.$row["Fname"].'</td><td>'.$row["Lname"].'</td><td>'.$row["email"].'</td><td>'.$row["contact"].'</td><td><a data-id="'.$row["Id"].'" class="delete" href="?id="'.$row["Id"].'>Delete</a></td></tr>';
    }
    echo "</table>";

和JS一样

<script>
    $(document).ready(function(){
      $(".delete").click(function(){
       var myId=$(this).data('Id');
       alert(myId);
          $.ajax({
             type:"Post",
             url:"delete.php",
             data:{
                 Id:myId,
              },
             sucess:function(data){

            }
            });
        });
    });
</script>

如何从使用jquery和ajax动态生成的表中获取id？

2 个答案: