我正在为我在学校的学生建立一个网站,这将使他们想要学习并获得xp积分,升级......但是当他们升级时,他们会得到一个选择的角色,这在我的数据库中是已知的作为armer_students。我想要的是一个包含sa_id(学生ID)和armer_id的更新系统,系统将更新armer_id。 一切都在我的页面上正确打印,但是当我点击更新按钮上的更新并刷新页面时,没有任何改变......
<?php
include 'db_functions.php';
$rows = db_select("SELECT * FROM armer_students");
if($rows === false) {
$error = db_error();
// Handle error - inform administrator, log to file, show error page, etc.
}
if(isset($_POST['update'])) {
// Quote and escape form submitted values
$sa_id = $_POST['sa_id'];
$armer_id = $_POST['armer_id'];
// Insert the values into the database
$sql = "UPDATE students SET armer_id= ".$armer_id ." WHERE sa_id=".$sa_id;
$result = db_query($sql);
}
?>
<html>
<head>
<title></title>
<link rel="stylesheet" type="text/css" href="style.css" >
</head>
<body>
<?php include 'menu.php'; ?>
<table>
<thead>
<tr>
<th>Id</th>
<th>armer id</th>
</tr>
</thead>
<?php foreach($rows as $row) :?>
<tr>
<td><?php echo $row['sa_id'] ;?></td>
<td><?php echo $row['armer_id'] ;?></td>
</tr>
<?php endforeach ;?>
</table>
<hr />
<tr>
<td width = "100">sa_id</td>
<td><input name = "sa_id" type = "number" id = "sa_id"></td>
</tr>
<tr>
<td width = "100">armer_id</td>
<td><input name = "armer_id" type = "number" id = "armer_id"></td>
</tr>
<tr>
<td width = "100"> </td>
<td> </td>
</tr>
<tr>
<td width = "100"> </td>
<td>
<input name = "update" type = "submit" id = "update" value = "Update">
</td>
</tr>
</table>
</form>
</body>
</html>
非常感谢:-)