我的目标是使用更新查询更新数据库中的值。在我的第一页上，我刚刚在网页中显示了数据库表。然后通过使用超链接我必须点击编辑到第二页“edit.php”。在第一页上我必须得到 id的值并将其发送到第二页。显示输入表单，其中随意获取值但 Id 通过隐藏标记。在第三页上实现了值查询，但缺少 id 的值。
<html>
<head>
<title>Assignment</title>
</head>
<body>
<?php
$con=mysql_connect("localhost","root","");
// Check connection
if (!mysql_connect()) {
echo "Failed to connect to MySQL: " . mysql_connect_error();
}
$db=mysql_select_db("assignment",$con);
$result = mysql_query("SELECT * FROM teacher ",$con);
?><table cellpadding="2px" border="2px"><?php
while($row = mysql_fetch_array($result)) {
?> <tr>
<td><a href="edit.php?id=<?php
echo $row['id']; ?>">Edit</a > <a href="delete.php">Delete</a>
</td><td>
<?php
echo $row['id']; ?></td><td> <?php echo $row['name'];?></td><td><?php echo $row['program']; ?></td>
<?php }
?></table><?php
mysql_close($con);
?>
</body>
</html>
<html>
<head>
<title>Assignment Edit</title>
</head>
<body>
<?php
$id = $_GET['id'];
?>
<form action="update.php" method="get">
Address <input type="text" name="program"><br>
<input type="hidden" name="id" value='<?php $id?>'>
<input type="submit" name="submit">
</form>
</body>
</html>
<html>
<head>
<title>Update Page</title>
</head>
<body>
<?php
$add=$_GET['program'];
$id=$_GET['id'];
$con=mysql_connect("localhost","root","");
// Check connection
if (!mysql_connect()) {
echo "Failed to connect to MySQL: " . mysql_connect_error();
}
$db=mysql_select_db("assignment",$con);
$query = "UPDATE teacher SET program='$add' WHERE id =".$id;
echo $query;
$result = mysql_query($query,$con);
/* while($row = mysql_fetch_array($result)) {
echo $row['id'] ." " . $row['name']." ". $row['address']."<br>";
}
mysql_close($con);
*/
?>
</body>
</html>
UPDATE老师SET program ='openSource'WHERE id =
答案 0 :(得分：0)
你需要改变这个
<input type="hidden" name="id" value='<?php $id?>'>
到
<input type="hidden" name="id" value='<?php echo $id?>'>
（或）
<input type="hidden" name="id" value='<?=$id?>'>