我正在构建一个列表，用户可以在其中更新或删除现有联系人。我创建了index.php，它成功地将联系人发送到数据库。 list.php 在表格中显示从index.php 输入的联系人列表。现在，用户应该删除或编辑每个联系人。
不幸的是，我的 edit_user.php 在点击“编辑”后返回错误: 您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，了解在第 1 行 '' 附近使用的正确语法。 另外，当我在 list.php 中点击“编辑”时，我希望 edit_user.php 显示带有预填充联系信息的编辑表单。
我是网络开发的新手。抱歉，代码太多了。请帮助我发现我的错误。 这是config.php
<?php
$dbhost = 'mysql51-031.wc2.dfw1.stabletransit.com';
$dbuser = '549359_sargis';
$dbpass = '********';
$dbname = '549359_sargis';
$table = 'Contacts';
$connection = mysql_connect($dbhost,$dbuser,$dbpass) or die(mysql_error());
$select_db = mysql_select_db($dbname,$connection) or die(mysql_error());
?>
这是我的 list.php
<?php
include("config.php");
?>
<html>
<head>
<title>Contact List</title>
<link rel="stylesheet" type="text/css" href="style.css" />
<a href="index.php">Create New Contact</a><hr/>
</head>
<body>
<?php
$result = mysql_query("SELECT * FROM Contacts", $connection);
$num_rows = mysql_num_rows($result);
if($num_rows > 0)
{
echo "<center><h1>Contact List: (Updated)</h1><table border = '1'>";
echo "<thead>";
echo "<tr>";
echo "<th> Firstname </th>";
echo "<th> Lastname </th>";
echo "<th> Email </th>";
echo "<th> Phone </th>";
echo "<th> Date </th>";
echo "<th> Action </th>";
echo "</th>";
echo "</thead>";
echo "<tbody>";
$query = mysql_query("SELECT * FROM Contacts");
while($record = mysql_fetch_array($query))
{
echo "<tr>";
echo "<td>" . $record['firstname'] . "</td>";
echo "<td>" . $record['lastname'] . "</td>";
echo "<td>" . $record['email'] . "</td>";
echo "<td>" . $record['phone_number'] . "</td>";
echo "<td>" . $record['timesstamp'] . "</td>";
echo "<td align='center'>"; ?>
<a href="edit_user.php"><input type="hidden" name="id" value="<?php echo $id; ?>" />Edit</a>
<?php echo "| <a href='list.php?action=delete&id=$id'>Delete</a></td>";
echo "</tr>";
}
echo "</table>";
echo "</center>";
}
else
echo "<center><h4>No contacts found.</h4></center>";
?>
</body>
</html>
这是 edit_user.php
<?php
include("config.php");
?>
<html>
<head>
<title>Edit User</title>
<link rel="stylesheet" type="text/css" href="style.css" />
</head>
<?php
if(isset($_POST['submit']))
//if (isset($_POST))
{
$id = intval($_POST['id']);
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$phonenumber = $_POST['phonenumber'];
$sql = "UPDATE Contacts SET firstname='".mysql_real_escape_string($firstname)."', lastname='".mysql_real_escape_string($lastname)."', email='".mysql_real_escape_string($email)."', phone_number='".mysql_real_escape_string($phonenumber)."', timesstamp =NOW() WHERE id=".mysql_real_escape_string($id);
//$sql = "UPDATE Contacts SET firstname='$firstname', lastname='$lastname', email='$email', phone_number='$phonenumber', timesstamp =NOW() WHERE id=$id";
//print_r($_POST).'<br />';echo $sql;exit;
$result = mysql_query($sql);
if($result)
{
header("Location: list.php");
}
else
{
echo "There was a problem with the query: ".mysql_error().".";
}
}
?>
<body>
<form action="edit_user.php?id=<?php echo $id;?>" method="POST">
<div>
First name: <input type ='text' id='firstname' name='firstname' value="<?php echo $firstname; ?>"/><br />
Last name: <input type = 'text' id='lastname' name='lastname'value="<?php echo $lastname; ?>"/><br />
Email: <input type = 'text' id='email' name='email' value="<?php echo $email; ?>"/><br />
Phone Number: <input type = 'text' id='phone_number' name='phonenumber' value="<?php echo $phonenumber; ?>"/><br />
<input type = 'submit' name = 'submit' value='Update' />
</div>
</form>
</body>
</html>
最佳答案
通常，此错误是由于 SQL 语句中的无效字符未转义造成的。
但我建议更改一些内容。
首先移动
$id = $_POST['id'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$phonenumber = $_POST['phonenumber'];
到之后
if(isset($_POST['submit'])) {
此外，为了好玩，请更改:
$id = $_POST['id'];
至:
$id = intval($_POST['id']);
然后在 SQL 语句中将其更改为:
$sql = "UPDATE Contacts SET firstname='".mysql_real_escape_string($firstname)."', lastname='".mysql_real_escape_string($lastname)."', email='".mysql_real_escape_string($email)."', phone_number='".mysql_real_escape_string($phonenumber)."', timesstamp =NOW() WHERE id=".mysql_real_escape_string($id);
关于php - 通过 mysql 更新 PHP 中的表单，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18581893/