如何在php中的bootstrap模式弹出窗口中显示从mysql数据库中选择的记录

Question

我想显示在模式弹出窗口中选择的特定记录，以便我可以编辑它，但记录不会显示在模态弹出窗口中。

我不知道如何传递ID href="#myModalDetail?id=<?=$productid?>。

以下是我正在使用的代码：

$query="SELECT * FROM product";
 $sql_q=mysql_query($query) or die(mysql_error());
  while($row = mysql_fetch_array($sql_q))
    {
    $productid=$row['pid'];
    $prodName= $row['product_name']; 
        }
<a class="btn btn-info" data-toggle="modal" href="#myModalDetail?id=<?=$productid?>"> <i class="fa fa-edit"></i> </a>


<div class="modal fade" id="myModalDetail">
   $productid=$_REQUEST['id'];
 $query="SELECT * FROM product WHERE pid='".$productid."'";
 $sql_q=mysql_query($query);

    while($row = mysql_fetch_array($sql_q))
    {
        $productid=$row['pid'];
        $prodName= $row['product_name']; 

    }
?>

?>

<div class="modal-dialog">
  <div class="modal-content">
    <div class="modal-header">
      <button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
      <h4 class="modal-title">Edit Products Details</h4>
    </div>
    <div class="modal-body">
      <div class="row">
        <div class="col-lg-12">
          <form role="form" name="Insertdb" method="post" action="edit-product.php">
            <div class="row">
              <div class="col-lg-4">
                <div class="form-group">
                  <label>Product Name</label>
                </div>
              </div>
              <div class="col-lg-6">
                <input class="form-control" name="prodName" value="<?=$prodName ?>">
              </div>
            </div>
          </div>
      </div>
    </div>
    <div class="modal-footer">
      <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
      <input name="button1" type="submit" class="btn btn-primary"> 
      </form>
    </div>
  </div>
  <!-- /.modal-content --> 
</div>
<!-- /.modal-dialog -->
</div>

</div>