【发布时间】：2013-02-14 05:54:17
【问题描述】：
有人可以帮帮我吗，我正在尝试提交超过 3 页的表格。每个都有 3 个文本区域字段，我使用 session start 来回显其他页面的表单数据。
所以最后我要做的就是回显表单数据并将其插入 mysql 表 ptb_registrations。
由于某种原因，尽管它无法正常工作并且我收到了更新数据库错误的错误。我已经为此工作了几个小时，我很抱歉地说，我想不通。请有人帮助我，告诉我哪里可能出错了。
第 1 页：
<?php
session_start();
?>
<form class="" method="post" action="register_p2.php">
<input type="text" id="first_name" name="first_name" placeholder="First Name" />
<input type="text" id="last_name" name="last_name" placeholder="Last Name" />
<input type="email" id="email" name="email" placeholder="Email" />
<input type="submit" value="Next >" />
</form>
第 2 页：
<?php
session_start();
// other php code here
$_SESSION['first_name'] = $first_name;
$_SESSION['last_name'] = $last_name;
$_SESSION['email'] = $email;
?>
<form name="myForm" method="post" action="register_p3.php" onsubmit="return validateForm()" >
<input type="text" id="date_of_birth" name="date_of_birth" placeholder="D.O.B 10/02/1990" />
<input type="text" id="number" name="number" placeholder="Mobile Number" />
<input type="text" id="confirm" name="confirm" placeholder="Are You a UK resident?" />
<input type="submit" value="Next >" />
</form>
第 3 页：
<?php
session_start();
// other php code here
$_SESSION['first_name'] = $first_name;
$_SESSION['last_name'] = $last_name;
$_SESSION['email'] = $email;
$_SESSION['dat_of_birth'] = $date_of_birth;
$_SESSION['number'] = $number;
?>
<form class="" method="post" action="register_p4.php">
<input type="text" id="display_name" name="date_of_birth" placeholder="Display Name" />
<input type="password" id="password" name="password" placeholder="Password" />
<input type="password" id="password2" name="password2" placeholder="Password (Confirm)" />
<input type="submit" value="Next >" />
</form>
第4页：（mysql函数）
<?php
session_start();
// other php code here
$_SESSION['first_name'] = $first_name;
$_SESSION['last_name'] = $last_name;
$_SESSION['email'] = $email;
$_SESSION['dat_of_birth'] = $date_of_birth;
$_SESSION['number'] = $number;
$_SESSION['display_name'] = $display_name;
$_SESSION['password'] = $password;
?>
<?php
////// SEND TO DATABASE
/////////////////////////////////////////////////////////
// Database Constants
define("DB_SERVER", "localhost");
define("DB_USER", "root");
define("DB_PASS", "");
define("DB_NAME", "database");
// 1. Create a database connection
$connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS);
if (!$connection) {
die("Database connection failed: " . mysql_error());
}
// 2. Select a database to use
$db_select = mysql_select_db(DB_NAME,$connection);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
//////////////////////////////////////////////////////////////
$query="INSERT INTO ptb_registrations (ID,
first_name,
last_name,
email,
date_of_birth,
contact_number,
display_name,
password
)
VALUES('NULL',
'".$first_name."',
'".$last_name."',
'".$email."',
'".$date_of_birth."',
'".$number."',
'".$display_name."',
'".$password."'
)";
mysql_query($query) or die ('Error updating database');
?>
<?php
function confirm_query($result_set) {
if (!$result_set) {
die("Database query failed: " . mysql_error());
}
}
function get_user_id() {
global $connection;
global $email;
$query = "SELECT *
FROM ptb_registrations
WHERE email = \"$email\"
";
$user_id_set = mysql_query($query, $connection);
confirm_query($user_id_set);
return $user_id_set;
}
?>
<?php
$user_id_set = get_user_id();
while ($user_id = mysql_fetch_array($user_id_set)) {
$cookie1 = "{$user_id["id"]}";
setcookie("ptb_registrations", $cookie1, time()+3600); /* expire in 1 hour */
}
?>
<?php include ('includes/send_email/reg_email.php'); ?>
<? ob_flush(); ?>
【问题讨论】：
-
通过 $_POST['First_Name'] 您必须在第二页中捕获该值，而不是将其称为变量
-
你是从哪里学来的？这看起来充满了SQL injection bugs，这会使将其放在互联网附近的任何地方都非常危险。
-
如果我不得不猜测数据库错误 - 不要在插入中提供 ID。取出ID和“null”。
-
使用 $_POST[] 而不是仅使用 $variabale_name。