Question

我输入用户名和密码时会收到此错误。

致命错误：在第32行的Z：\ xampp \ htdocs \ zed \ checklogin.php中调用未定义的函数session_register（）

这是代码。

main_login.php

<!DOCTYPE html>
</html lang="en">
    <head>
        <meta charset="utf-8">
        <title>Test</title>
    </head>
    </body> 
        <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
        <tr>
        <form name="form1" method="post" action="checklogin.php">
        <td>
        <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
        <tr>
        <td colspan="3"><strong>Member Login </strong></td>
        </tr>
        <tr>
        <td width="78">Username</td>
        <td width="6">:</td>
        <td width="294"><input name="myusername" type="text" id="myusername"></td>
        </tr>
        <tr>
        <td>Password</td>
        <td>:</td>
        <td><input name="mypassword" type="text" id="mypassword"></td>
        </tr>
        <tr>
        <td>&nbsp;</td>
        <td>&nbsp;</td>
        <td><input type="submit" name="Submit" value="Login"></td>
        </tr>
        </table>
        </td>
        </form>
        </tr>
        </table>
    </body>
</html>

checklogin

<?php

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="password"; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword']; 

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){

// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword"); 
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
?>

login_success.php

// Check if session is not registered, redirect back to main page. 
// Put this code in first line of web page. 
<?php
session_start();
if(!session_is_registered(myusername)){
header("location:main_login.php");
}
?>

<html>
<body>
Login Successful
</body>
</html>

logout.php

// Put this code in first line of web page. 
<?php 
session_start();
session_destroy();
?>

Answer 1

您正在关注一个非常古老的（^和错误的）教程。不推荐使用所有mysql_*函数，您应该查看PDO ¹或MySQLi ²

您的错误消息：

Fatal error: Call to undefined function session_register() in Z:\xampp\htdocs\zed\checklogin.php on line 32

因为session_register()也被弃用的国家/地区。您需要设置如下值：

$_SESSION['username'] = 'theusername';

然后在您检查用户是否已登录的页面上，您应该使用以下内容：

session_start();
if(!isset($_SESSION['username']) || empty($_SESSION['username'])) {
    die(header("Location: /login"));
}

¹ - PDO

² - MySQLi

对于额外的Cookie分数，这是一个示例PDO查询。

$db = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'username', 'password');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);

$stmt = $db->prepare("SELECT * FROM $tbl_name WHERE username=? and password=?");
$stmt->bindValue(1, $username, PDO::PARAM_STR);
$stmt->bindValue(2, $password, PDO::PARAM_STR);
if($stmt->execute()){
    if($stmt->rowCount() == 1) {
        $data = $stmt->fetchObject();
        $_SESSION['username']
    }
}

^{免责声明：以上是伪代码，我建议您查看PDO PHP示例。}

Php登录错误

1 个答案: