我正在创建一个登录页面,用户和管理员将在其中登录用户将具有角色 = 用户和状态 = 待定,直到管理员将其激活。我有不同的文件要显示给用户和管理员,在用户中,有 2 个文件。1 个用于活动用户,另一个用于待定用户。
我创建了 if 语句并尝试了 switch 语句。但我在 XAMPP 上遇到错误“解析错误:语法错误,C:\xampp\htdocs\MakerLab\server.php 中第 109 行的文件意外结束”
这是我的 server.php
...
<?php
session_start();
// variable declaration
$email = "";
$status = "";
$errors = array();
$_SESSION['success'] = "";
// connect to database
$db = mysqli_connect('localhost', 'root', '', 'makerlab');
// REGISTER USER
if (isset($_POST['reg_user'])) {
// receive all input values from the form
$fname = mysqli_real_escape_string($db, $_POST['fname']);
$lname = mysqli_real_escape_string($db, $_POST['lname']);
$email = mysqli_real_escape_string($db, $_POST['email']);
$lewisID = mysqli_real_escape_string($db, $_POST['lewisID']);
$password_1 = mysqli_real_escape_string($db, $_POST['password_1']);
$password_2 = mysqli_real_escape_string($db, $_POST['password_2']);
// form validation: ensure that the form is correctly filled
//if (empty($email)) { array_push($errors, "Lewis Email is required"); }
//if (empty($password_1)) { array_push($errors, "Password is required"); }
//if ($password_1 != $password_2) {
// array_push($errors, "The two passwords do not match");
//}
$user_check_query = "SELECT * FROM users WHERE lewisID='$lewisID' OR email='$email' LIMIT 1";
$result = mysqli_query($db, $user_check_query);
$user = mysqli_fetch_assoc($result);
if ($user) { // if user exists
if ($user['lewisID'] === $lewisID) {
array_push($errors, "lewisID already exists");
}
if ($user['email'] === $email) {
array_push($errors, "lewisID already exists");
}
}
// register user if there are no errors in the form
if (count($errors) == 0) {
$password = md5($password_1);//encrypt the password before saving in the database
$query = "INSERT INTO users (lewisID,
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