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我在$_SESSION變量中遇到了很多麻煩。我正在嘗試爲用戶登錄和註銷創建一種方法。我可以登錄一個用戶，但我似乎無法保持會話，當我切換頁面。當用戶正確登錄時，他們被帶到profile.php。但是，如果我回到index.php打印以下錯誤：無法驗證用戶身份
Notice: Undefined index: login in /Applications/MAMP/htdocs/www/Shared sites/userlogreg/index.php on line 3
我很新的這一點，但距離使面色上和其他地方我似乎無法弄清楚。任何幫助，將不勝感激。
的index.php
<?php
session_start();
if ($_SESSION['login'] == 1) {
echo "<h1>Logged in!</h1>";
} else {
echo "<h1>Not logged in</h1><br/>";
}
?>
<!DOCTYPE HTML>
<html>
<head>
<title>Index page</title>
</head>
<body>
<h2>Login</h2>
<form action="login.php" method="POST">
<div>
<label for="emailSignIn">Email:</label>
<input type="email" name="email" placeholder="Email" required="required" />
</div>
<div>
<label for="passwordSignIn">Password:</label>
<input type="password" name="password" placeholder="Password" required="required" />
</div>
<input type="submit" name="submit" value="Sign in" />
</form>
<h2>Register</h2>
<form action="register.php" method="POST">
<div>
<label for="firstnameRegister">First name:</label>
<input type="text" name="firstname" placeholder="First name" required="required" />
</div>
<div>
<label for="lastnameRegister">Last name:</label>
<input type="text" name="lastname" placeholder="Last name" required="required" />
</div>
<div>
<label for="emailRegister">Email:</label>
<input type="email" name="email" placeholder="Email" required="required" />
</div>
<div>
<label for="passwordRegister">Password:</label>
<input type="password" name="password" placeholder="Password" required="required">
</div>
<input type="submit" name="submit" value="Create account" />
</form>
</body>
</html>
的login.php
<?php
$email = sanitize_input($_POST['email']); //echo "Sanitized email: ".$email; echo "<br/>";
$password = $_POST['password']; //echo "Inputted password: ".$password; echo "<br/>";
if ((!isset($email)) || (!isset($password))) {
// VISITOR NEEDS TO ENTER AN EMAIL AND PASSWORD
//echo "Data not provided";
} else {
// CONNECT TO MYSQL
$mysql = mysqli_connect("localhost", "root", "root");
if(!$mysql) {
//echo "Cannot connect to PHPMyAdmin.";
exit;
} else {
}
}
// SELECT THE APPROPRIATE DATABASE
$selected = mysqli_select_db($mysql, "languageapp");
if(!$selected) {
//echo "Cannot select database.";
exit;
} else {
}
// GET THE USER'S UNIQUE SALT FROM THE DATABASE
$unique_salt = mysqli_query($mysql, "select uniqueSalt from user where email = '".$email."'");
$row = mysqli_fetch_array($unique_salt);
//echo "Salt: ".$row['uniqueSalt']; echo "<br/>";
// HASH THE PASSWORD
$iterations = 10;
$hashed_password = crypt($password,$row['uniqueSalt']);
for ($i = 0; $i < $iterations; ++$i)
{
$hashed_password = crypt($hashed_password . $password,$row['uniqueSalt']);
}
//echo "Password entered by user: ".$hashed_password; echo "<br/>";
$user_db_password = mysqli_query($mysql, "select password from user where email = '".$email."'");
$row = mysqli_fetch_array($user_db_password);
//echo "User's password: ".$row['password']; echo "<br/>";
// query the database to see if there is a record which matches
$query = "select count(*) from user where email = '".$email."' and password = '".$hashed_password."'";
$result = mysqli_query($mysql, $query);
if(!$result) {
//echo "Cannot run query.";
exit;
}
$row = mysqli_fetch_row($result);
$count = $row[0];
if ($count > 0) {
session_start();
$_SESSION['login'] = 1;
$_SESSION['email'] = $email;
$_SESSION['errors'] = "";
header("location:profile.php");
//echo "<h1>Login successful!</h1>";
//echo "<p>Welcome.</p>";
//echo "<p>This page is only visible when the correct details are provided.</p>";
} else {
session_start();
$_SESSION['login'] = '';
header("location:index.php");
//echo "<h1>Login unsuccessful!</h1>";
//echo "<p>The email and password combination entered was not recognized</p>";
}
// CLEAN THE INPUT
function sanitize_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
幾周前，我在尋找相同的東西時也遇到了麻煩，但最終在別人的博客上發現了它。 – 2012-08-15 18:01:13
你正在使用'mysqli'在正確的軌道上，但是你錯用了它。 'sanitize_input'是** NOT **在你的SQL查詢中使用適當的佔位符的替代。你應該**不要**直接使用'$'vars，而是使用'？'佔位符。 – tadman 2012-08-15 19:19:46
@tadman你能解釋一下嗎？這個想法是不是改變用戶的輸入，而是改變數據的重複（佔位符）？也許你可以指示我一些關於它的信息。謝謝。 – garethdn 2012-08-16 12:15:53