Question

我正在创建一个链接到数据库的网站（学校项目，新手在这里！），这将允许一个人登录和访问信息，但无论我做什么改变它，我都会继续提出这个错误。< / p>

解析错误：第38行/ Applications / XAMPP / xamppfiles / htdocs / Unit1 / Databases / Band Project / passCheck.php中的语法错误，意外的'$ userID'（T_VARIABLE）

我的passcheck.php代码如下：

<p>
  <?php

    $email = $_POST["email"];
    echo $email;
    echo "</br>";

    $password = $_POST["password"];
    echo $password;
    echo "</br>";

    require_once 'config.php';

    require_once 'databaseclass.php';

    $db = new databaseclass($pdo);


    $sql = "SELECT * FROM User WHERE email = '" . $email . "' AND password = '" . $password . "'";
    echo $sql;
    echo "<br>";

    $rows = $db->query($sql)->fetchAll();
    $count = count($rows);
    print_r($rows);
    echo "<br>";
    echo $count;

    if($count == 1 ){
        session_start();
            $_SESSION['firstname'] = $firstname;
            $_SESSION["lastname"] = $lastname;
            $_SESSION['userID'] = $userID;
            $_SESSION['loggedin'] = true;
            header('Location:http://localhost/Unit1/Databases/Band%20Project/homepage.php');}
    else{

        session_destroy();
        $error = "Your Login Name or Password is invalid";
        header('Location:http://localhost/Unit1/Databases/Band%20Project/login.html');
        echo $error;}

  ?>      
</p>

我的配置页面是：

<?php

$host = '127.0.0.1';
$db = 'bandProject';
$user = 'root';
$pass = '';
$charset = 'utf8';

try{wqcd
        $dsn = "mysql:host=$host;dbname=$db;charset=$charset";
        $opt = [
            PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
            PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_OBJ,
            PDO::ATTR_EMULATE_PREPARES   => false,
        ];

    $pdo = new PDO($dsn, $user, $pass, $opt);
}
catch (PDOException $e)
{
    exit('Error connecting to DataBase');

}

我的会话php是：

<?php
require_once 'config.php';
session_start();

$user_check = $_SESSION['userID'];


$ses_sql = mysqli_query($db,"select username from admin where username = '$user_check' ");

$row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);

$login_session = $row['username'];

if(!isset($_SESSION['login_user'])){
       header("location:http://localhost/Unit1/Databases/Band%20Project/login.html");
}
?>

我的主页是：

<h1><center><big><font color=#653C5E>Homepage</font></big></center></h1>
    <BR>
        <?php
        require_once 'session.php' ;
        if($count == 1 ){
        session_start();
            $_SESSION['firstname'] = $firstname;
            $_SESSION['lastname'] = $lastname;
            $_SESSION['userID'] = $userID;
            $_SESSION['loggedin'] = true;
            header('Location:http://localhost/Unit1/Databases/Band%20Project/homepage.php')
            echo "Hello " . $firstname . " " . $lastname;}
            //set all your session variables (firstname, lastname, userID)`
            //should have from SQL results
            //open homepage//
        else{

            session_destroy();
            $error = "Your Login Name or Password is invalid";
            echo $error;
            header('Location:http://localhost/Unit1/Databases/Band%20Project/login.html');}

        ?>

    <BR>
            </BODY>

Answer 1

尝试;

$_SESSION['userID'] = $rows["userID"];

解析错误：语法错误，第38行意外的'$ userID'（T_VARIABLE）

1 个答案: