PHP - 将隐藏值传递给jquery [英] PHP - pass hidden value into the jquery
问题描述
< HTML>
< head>
< link rel =stylesheethref =js / jquery-ui-themes-1.11.1 / themes / smoothness / jquery-ui.css/>
< script type =text / javascriptsrc =js / jquery-1.11.1.js>< / script>
< script type =text / javascriptsrc =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连接
$ 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 =600border =1cellpadding =0cellspacing =0style =border-collapse：collapse>
< tr>
< td width =80align =center> ID< / td>
< td width =300align =center>部门< / td>
< td width =220align =center>动作< / 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>删除< / button>
< input type =hiddenid =departmentIDvalue =<？php echo $ row ['dept_ID'];？> />
< / td>
< / tr>
<？php
}
？>
< / table>
<？php
}
？>
部门表格
dept_ID ; dept_name
1 Account
2 ;财务
3 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;营销
假设我的部门表只有3条记录。
我的要求如下：
- 点击第一个删除按钮，显示部门ID = 1
- 点击第二个删除按钮，显示部门ID = 2
- 点击第三个删除按钮，显示部门ID = 3
但是，从我的代码中，我无法满足我的要求。
我得到的部门ID输出是1，无论我点击哪个按钮。
有人可以帮我吗？
不需要使用隐藏输入，只需使用按钮标记：
<？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 =submitname =departmentIDclass =buttonsPromptConfirmDeleteDepartmentvalue =<？php echo $ row ['dept_ID'];？>>删除< / 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'];
//在它旁边的一些进程
注意：不要忘记标记。
其他注意：不要忘记使用准备好的语句：
$ sql ='删除部门WHERE dept_ID =？';
$ stmt = $ dbc-> prepare（$ sql）;
$ stmt-> bind_param（'i'，$ id）;
$ stmt-> execute（）;
//一些想法，必要时使用错误检查
// $ dbc->错误
<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 $row['dept_ID']; ?>" />
</td>
</tr>
<?php
}
?>
</table>
<?php
}
?>
department table
dept_ID dept_name
1 Account
2 Finance
3 Marketing
Assume that my department table only have 3 records.
My requirement is the following:
- Click 1st delete button, show department ID = 1
- Click 2nd delete button, show department ID = 2
- Click 3rd delete button, show department ID = 3
However from my code, I can't meet my requirement.
The department ID output that I get is 1 no matter what button I clicked.
Can someone help me?
No need to use a hidden input, you could just use the button tag instead:
<?php while($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 } ?>
Of course, in the PHP script that does the form processing, access the POST index like you normally would:
$id = $_POST['departmentID'];
// some processes next to it
Note: Don't forget the <form> tag.
Additional Note: Don't forget to use prepared statements:
$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
这篇关于PHP - 将隐藏值传递给jquery的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
