2
有人可以幫助我，我試圖提交一個3頁以上的表格。在每個使用會話中的3個文本區域字段開始回顯其他頁面的表單數據。會話表單數據跨多個頁面並插入到mysql？
那麼最後我所要做的就是回顯表單數據並將其插入到mysql表中的ptb_registrations中。
由於某種原因，雖然它不工作，即時通訊錯誤更新數據庫錯誤。我一直在這個工作幾個小時，很抱歉說，我無法弄清楚。請有人幫助我，讓我看看我可能會出錯的地方。
頁面1：
<?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>
頁2：
<?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>
頁3：
<?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>
頁4：（MySQL的功能）
<?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", $cookie1, time()+3600); /* expire in 1 hour */
}
?>
<?php include ('includes/send_email/reg_email.php'); ?>
<? ob_flush(); ?>
通過$ _POST ['First_Name']，您必須捕獲第二頁的值而不是將其作爲變量引用 – 2013-02-14 05:57:18
您從哪裏學會了這樣做？這看起來像是充滿了[SQL注入漏洞]（http://bobby-tables.com/），如果把它放在互聯網附近的任何地方都會使它非常危險。 – tadman 2013-02-14 05:59:38
如果我不得不猜測數據庫錯誤 - 不要在插入中提供ID。取出ID和「null」。 – 2013-02-14 06:00:02