Question

<html>
<head>
<link rel="stylesheet" href="js/jquery-ui-themes-1.11.1/themes/smoothness/jquery-ui.css" />
<script type="text/javascript" src="js/jquery-1.11.1.js"></script>
<script type="text/javascript" src="js/jquery-ui-1.11.1/jquery-ui.js"></script>
<script>
$(document).ready(function(){
    $(".buttonsPromptConfirmDeleteDepartment").click(function(){
        var departmentID = $('input#departmentID').val();
        alert(departmentID);
    });
});
</script>
</head>

<body>
<?php
//db connection

$query = "SELECT * 
          FROM department 
          ORDER BY dept_ID ASC";
$result = mysqli_query($dbc, $query);
$total_department = mysqli_num_rows($result);

if($total_department > 0)
{
?>

<table width="600" border="1" cellpadding="0" cellspacing="0" style="border-collapse:collapse">
    <tr>
      <td width="80" align="center">ID</td>
      <td width="300" align="center">Department</td>
      <td width="220" align="center">Action</td>
    </tr>   
<?php        
    while($row = mysqli_fetch_array($result))
    {
?>
      <tr>
        <td align="center"><?php echo $row['dept_ID']; ?></td>
        <td align="center"><?php echo $row['dept_name']; ?></td>
        <td>
          <button class="buttonsPromptConfirmDeleteDepartment">Delete</button>
          <input type="hidden" id="departmentID" value="<?php echo $row['dept_ID']; ?>" />    
        </td>
      </tr>
<?php
   }
?>
  </table> 
<?php
}
?>

部门表
DEPT_ID DEPT_NAME
1个帐户
2财经
3个营销

假设我的部门表只有3条记录我的要求如下：
- 单击第一个删除按钮，显示部门ID = 1
- 单击第二个删除按钮，显示部门ID = 2
- 单击第3个删除按钮，显示部门ID = 3

但是从我的代码来看，我无法满足我的要求。
无论我点击什么按钮，我得到的部门ID输出都是1。

有人可以帮助我吗？

Answer 1

无需使用隐藏输入，您只需使用button标记：

<?php while($row = mysqli_fetch_array($result)) { ?>
    <tr>
        <td align="center"><?php echo $row['dept_ID']; ?></td>
        <td align="center"><?php echo $row['dept_name']; ?></td>
        <td>
            <button type="submit" name="departmentID" class="buttonsPromptConfirmDeleteDepartment" value="<?php echo $row['dept_ID']; ?>">Delete</button>
        </td>
    </tr>
<?php } ?>

当然，在执行表单处理的PHP脚本中，像通常那样访问POST索引：

$id = $_POST['departmentID'];
// some processes next to it

注意：不要忘记<form>标记。

附加说明：不要忘记使用预备语句：

$sql = 'DELETE FROM department WHERE dept_ID = ?';
$stmt = $dbc->prepare($sql);
$stmt->bind_param('i', $id);
$stmt->execute();
// some idea, use error checking when necessary
// $dbc->error

Answer 2

更改

ID = “DepartmentID的”

到

class =“departmentID”和

更改

<script>
$(document).ready(function(){
   $(".buttonsPromptConfirmDeleteDepartment").click(function(){
    var departmentID = $('input#departmentID').val();
    alert(departmentID);
   });
});

到

 <script>
 $(document).ready(function(){
    $(".buttonsPromptConfirmDeleteDepartment").click(function(){
    var departmentID = $(this).next('input.departmentID').val();
    alert(departmentID);
 });
 });

Answer 3

首先在while循环中使用dept_id并且你为所有dept使用相同的id .. 另一件事你可以使用jquery按钮点击dept_id ..像这样

$('.buttonsPromptConfirmDeleteDepartment').click(function(){
dept_id = $(this).next('input').val();
})

PHP - 将隐藏值传递给jquery

3 个答案: