【发布时间】:2011-08-28 10:44:13
【问题描述】:
我在 HTML 中创建了一个表单(查询表单),该表单发布到以下代码:
<?php
if(isset($_POST['submit']))
{
$name = mysql_real_escape_string((string)$_POST['name']);
$surname = mysql_real_escape_string((string)$_POST['surname']);
$email = mysql_real_escape_string((string)$_POST['email']);
$phone = mysql_real_escape_string((string)$_POST['phone']);
$country = mysql_real_escape_string((string)$_POST['country']);
$message = mysql_real_escape_string((string)$_POST['message']);
$sql = "INSERT INTO contact
(name, surname, email, phone, country, message)
VALUES('$name', '$surname', '$email', '$phone', '$country', '$message')";
mysql_select_db($db);
$retval = mysql_query( $sql, $conn )or die(mysql_error());
echo 'Thank you '.$name.' '.$surname.'. Your enquiry has been forwarded to our team. <br><br>Please check you email inbox for further information.<br><br>Return to homepage:<br><br><button class="search" onclick="/">Return to homepage</button>';
mysql_close($conn);
}
?>
我想知道,当输入无效或零数据时,如何显示错误并停止表单发布?
在学习如何在 Web 上创建表单的同时,我还听说过 SQL 注入。我受到保护了吗?
非常感谢您的帮助。
【问题讨论】:
标签: php mysql forms validation