登錄系統中的未定義變量會話
[英]Undefined variable session in a login system
問題描述
嗨，我正在嘗試開發一個登錄系統，但是在某種情況下我似乎遇到了錯誤。這是我的代碼：
//header.php
if($_SESSION['signed_in']){
echo 'Hello' . $_SESSION['user_name'] . '. Not you? <a href="signout.php">Sign out</a>';
}
else{
echo '<a href="signin.php">Sign in</a> or <a href="sign up">create an account</a>.';
}
//Signin.php
if(mysql_num_rows($result) == 0)
{
echo 'You have supplied a wrong user/password combination. Please try again.';
}
else
{
//set the $_SESSION['signed_in'] variable to TRUE
$_SESSION['signed_in'] = true;
//we also put the user_id and user_name values in the $_SESSION, so we can use it at various pages
while($row = mysql_fetch_assoc($result))
{
$_SESSION['user_id'] = $row['user_id'];
$_SESSION['user_name'] = $row['user_name'];
$_SESSION['user_level'] = $row['user_level'];
}
echo 'Welcome, ' . $_SESSION['user_name'] . '. <a href="index.php">Proceed to the forum overview</a>.';
}
該錯誤指向第一個條件：if（$ _ SESSION ['signed_in']），它表示：
注意：未定義的變量：第21行的C：\\ xampp \\ htdocs \\ Tutorials \\ Forum \\ header.php中的_SESSION
我如何才能做到這一點？
編輯 ：session_start（）包含在doctype的header.php文件的頂部，而header.php包含在Signin.php中
完整代碼：
header.php
<?php
session_start();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="nl" lang="nl">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<meta name="description" content="A short description." />
<meta name="keywords" content="put, keywords, here" />
<title>PHP-MySQL forum</title>
<link rel="stylesheet" href="css/style.css" type="text/css">
</head>
<body>
<h1>My forum</h1>
<div id="wrapper">
<div id="menu">
<a class="item" href="/forum/index.php">Home</a> -
<a class="item" href="/forum/create_topic.php">Create a topic</a> -
<a class="item" href="/forum/create_cat.php">Create a category</a>
<div id="userbar">
<div id="userbar">
<?php
if($_SESSION['signed_in']){
echo 'Hello' . $_SESSION['user_name'] . '. Not you? <a href="signout.php">Sign out</a>';
}
else{
echo '<a href="signin.php">Sign in</a> or <a href="sign up">create an account</a>.';
}
?>
</div>
</div>
<div id="content">
Signin.php
<?php
//signin.php
include 'conn.php';
include 'header.php';
echo '<h3>Sign in</h3>';
//first, check if the user is already signed in. If that is the case, there is no need to display this page
if(isset($_SESSION['signed_in']) && $_SESSION['signed_in'] == true)
{
echo 'You are already signed in, you can <a href="signout.php">sign out</a> if you want.';
}
else
{
if($_SERVER['REQUEST_METHOD'] != 'POST')
{
/*the form hasn't been posted yet, display it
note that the action="" will cause the form to post to the same page it is on */
echo '<form method="post" action="">
Username: <input type="text" name="user_name" />
Password: <input type="password" name="user_pass">
<input type="submit" value="Sign in" />
</form>';
}
else
{
/* so, the form has been posted, we'll process the data in three steps:
1. Check the data
2. Let the user refill the wrong fields (if necessary)
3. Varify if the data is correct and return the correct response
*/
$errors = array(); /* declare the array for later use */
if(!isset($_POST['user_name']))
{
$errors[] = 'The username field must not be empty.';
}
if(!isset($_POST['user_pass']))
{
$errors[] = 'The password field must not be empty.';
}
if(!empty($errors)) /*check for an empty array, if there are errors, they're in this array (note the ! operator)*/
{
echo 'Uh-oh.. a couple of fields are not filled in correctly..';
echo '<ul>';
foreach($errors as $key => $value) /* walk through the array so all the errors get displayed */
{
echo '<li>' . $value . '</li>'; /* this generates a nice error list */
}
echo '</ul>';
}
else
{
//the form has been posted without errors, so save it
//notice the use of mysql_real_escape_string, keep everything safe!
//also notice the sha1 function which hashes the password
$sql = "SELECT
user_id,
user_name,
user_level
FROM
users
WHERE
user_name = '" . mysql_real_escape_string($_POST['user_name']) . "'
AND
user_pass = '" . sha1($_POST['user_pass']) . "'";
$result = mysql_query($sql);
if(!$result)
{
//something went wrong, display the error
echo 'Something went wrong while signing in. Please try again later.';
//echo mysql_error(); //debugging purposes, uncomment when needed
}
else
{
//the query was successfully executed, there are 2 possibilities
//1. the query returned data, the user can be signed in
//2. the query returned an empty result set, the credentials were wrong
if(mysql_num_rows($result) == 0)
{
echo 'You have supplied a wrong user/password combination. Please try again.';
}
else
{
//set the $_SESSION['signed_in'] variable to TRUE
$_SESSION['signed_in'] = true;
//we also put the user_id and user_name values in the $_SESSION, so we can use it at various pages
while($row = mysql_fetch_assoc($result))
{
$_SESSION['user_id'] = $row['user_id'];
$_SESSION['user_name'] = $row['user_name'];
$_SESSION['user_level'] = $row['user_level'];
}
echo 'Welcome, ' . $_SESSION['user_name'] . '. <a href="index.php">Proceed to the forum overview</a>.';
}
}
}
}
}
include 'footer.php';
?>
2 個解決方案
解決方案1
2 2012-10-17 09:49:47
沒有session_start(); 這就是原因，您必須始終使用PHP中的會話聲明該befoe。
解決方案2
2 已采納 2012-10-17 11:09:57
在將內容存儲到會話中之前，變量$_SESSION['signed_in']將為空。 發生警告是因為您要求此值，但里面什么也沒有。
if (isset($_SESSION['signed_in']) && $_SESSION['signed_in'])
{
}
為了避免警告，您應該首先檢查變量是否存在，然后才能從中讀取。 當然，這有點麻煩，因此大多數開發人員都創建了一個僅用於從數組安全讀取的函數。
編輯：
實際上，上述問題會導致另一條消息...
Notice: Undefined index: ...
就像Mob所說的那樣，變量$ _SESSION根本不存在，因為沒有會話開始。 我將保留這個答案，因為這將是下一個陷阱。
php symfony 4 FosUserBundle 無法使用防火牆模式登錄 ^/login & 未定義變量 _SESSION
[英]php symfony 4 FosUserBundle cannot login with firewall pattern ^/login & undefined variable _SESSION
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