你好，我正在尝试开发一个登录系统，但我似乎在某个条件下遇到错误。这是我的代码:
//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, ' . $_SESSION['user_name'] . '. <a href="index.php">Proceed to the forum overview</a>.';
}
错误指向第一个条件:if($_SESSION['signed_in']) 它说:
注意: undefined variable :C:\xampp\htdocs\Tutorials\Forum\header.php 第21行的_SESSION
我该如何纠正？
编辑:session_start() 包含在文档类型中的 header.php 文件的顶部，而 header.php 包含在 Signin.php 中
完整代码:
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';
?>
最佳答案
在将某些内容存储到 session 中之前，变量 $_SESSION['signed_in'] 将为空。出现警告是因为您要求提供此值，但里面没有任何内容。
if (isset($_SESSION['signed_in']) && $_SESSION['signed_in'])
{
}
为避免警告，您应该首先检查变量是否存在，然后您可以从中读取。当然这有点麻烦，所以大多数开发人员创建一个函数只是为了从数组中安全读取。
编辑:
实际上上面的问题会导致另一个消息...
Notice: Undefined index: ...
...所以正如Mob所说，变量$_SESSION根本不存在，因为没有启动 session 。我会让这个答案留下来，因为这将是下一个陷阱。
关于php - 登录系统中 undefined variable session ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/12931503/