php - 将 mysql 的值传递给 bootstrap modal

Question

我想从数据列表中进行选择并在模式弹出窗口中显示所选数据的结果。我只是不知道如何将 id 传递给弹出窗口，以便我可以显示结果。帮助!!

<td> <button class="btn btn-primary " data-toggle="modal" data-id="<?php $id= $row['id'] ?>" data-target="#myModal"  <?php echo"id=$row[id]'";?> >
          View Details
       </button>
 </td>

模式弹出窗口

<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
      <div class="modal-dialog">
          <div class="modal-content">
              <div class="modal-header">
                  <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>
                      <h4 class="modal-title" id="myModalLabel">Requisition Details</h4>
              </div>
              <div class="modal-body">
                  <?php
                        $id = $_GET['id'];
                        include('mysql_connect.php');
                        $query11 =  mysql_query ("select * from p_requisition where id = '$id' ") or die(mysql_error());
                        $rows= mysql_fetch_array($query11);
                        echo $rows['details'];
                  ?>                                       
             </div>
             <div class="modal-footer">
                 <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
                 <button type="button" class="btn btn-primary">Save changes</button>
              </div>
           </div>
                                    <!-- /.modal-content -->
         </div>
                                <!-- /.modal-dialog -->
    </div>

Answer 1

首先，为按钮添加类可能是get-data，

然后在jquery中，

$(document).ready(function() {
    $(document).on("click",".get-data"function(){
       var val = $(this).attr("data-id");    
       $.ajax({
            url: "path to ajax file",
            type: "POST",
            dataType: "HTML",
            async: false,
            success: function(data) {
               $('.modal-body').html(data);          
           }
         }); 

     });
});

在ajax文件中，编写查询从mysql获取数据，操作数据并在弹出窗口中显示。

ajax.php

<?php
     $id = $_POST['id'];
     include('mysql_connect.php');
     $query11 =  mysql_query ("select * from p_requisition where id = '$id' ") or die(mysql_error());
     $rows= mysql_fetch_array($query11);
     echo $rows['details'];// here you need to manipulate the database data with html and pass it on to popup.
?>