警告：mysqli_stmt_bind_param()：類型定義字符串中的元素數與中的綁定變量數不匹配

Question

我正在用 Php 和 MySqli 做一個登錄系統，我剛剛完成了注冊功能。

設置 error_reporting(E_ALL) 后，我得到了上述錯誤，我得到的兩個錯誤之一是即使我的 if 語句沒有通過輸入，直到被發送到數據庫。 這是我的代碼：

<?php

require_once "partials/header.php";
error_reporting(E_ALL);

if (isset($_POST["signup-button"])) {
require "includes/dbconn.php";

$username = filter_var($_POST["username"], FILTER_SANITIZE_STRING);
$username = trim($username);

$email = filter_var($_POST["email"], FILTER_VALIDATE_EMAIL);
$email = trim($email);

$password = filter_var($_POST["password"], FILTER_SANITIZE_STRING);
$passwordRe = filter_var($_POST["passwordRe"], FILTER_SANITIZE_STRING);

$termsOfService = filter_var($_POST["terms"], FILTER_SANITIZE_STRING);
$errors = [];

if (strlen($username) < 5) {
    $errors[] = "Your Username should contain at least 5 characters";
}

if (!$email) {
    $errors[] = "Your E-Mail is invalid.";
}

if (strlen($password) < 6) {
    $errors[] = "Your password should contain at least 6 characters.";
}

if ($password !== $passwordRe) {
    $errors[] = "Both passwords should be identical";
}

if (!$termsOfService) {
    $errors[] = "Please accept our Terms of Service";
} else {
    /*$sql = "INSERT INTO usersvet (userName, email, password) VALUES
                    ('" . $username . "', '" . $email . "', '" . $hashedPwd . "')";

        if (!mysqli_query($dbConnection, $sql)) {
            die(mysqli_error($dbConnection));*/
    $sql = "SELECT userName FROM usersvet WHERE userName=? AND pwd=?";
    $statement = mysqli_stmt_init($dbConnection);

        if (!mysqli_stmt_prepare($statement, $sql)) {
            $errors[] = "Sql Error.";
        } else {
            mysqli_stmt_bind_param($statement, "s", $username);
            mysqli_stmt_execute($statement);
            mysqli_stmt_store_result($statement);
            $resultCheck = mysqli_stmt_num_rows($statement);

            if ($resultCheck > 0) {
                $errors[] = "User already taken.";                 
            } else {
                $sql = "INSERT INTO usersvet (userName, email, pwd)
                                            VALUES (?, ?, ?);";
                $statement = mysqli_stmt_init($dbConnection);

                if (!mysqli_stmt_prepare($statement, $sql)) {
                    $errors[] = "SQL Error.";
                } else {
                    $hashedPwd = password_hash($password, PASSWORD_DEFAULT);
                    mysqli_stmt_bind_param($statement, "sss", $username, $email, $hashedPwd);
                    mysqli_stmt_execute($statement);
                    mysqli_stmt_store_result($statement);
                }
            }
        }
    }
    mysqli_stmt_close($statement);
    mysqli_close($dbConnection);
}

前面提到的錯誤數組用於以 div 的形式向頁面發送視覺反饋，代碼如下：

<?php if (!empty($errors)): ?>
    <?php foreach ($errors as $error): ?>
        <div class="well">
            <p class="alert alert-warning"><?= $error ?></p>
        </div>
    <?php endforeach ?>
<?php endif ?>

<?php if (!empty($_POST) && empty($errors)): ?>
        <div class="well">
            <p class="alert alert-success">Sign Up successful. You can now login <a href="login.php">here</a>.</p>
        </div>
<?php endif ?>

Answer 1

如果您的目標是確定用戶名是否已存在，則不必關心密碼。

你有這個：

$sql = "SELECT userName FROM usersvet WHERE userName=? AND pwd=?";
mysqli_stmt_bind_param($statement, "s", $username);

你要這個：

$sql = "SELECT userName FROM usersvet WHERE userName=?";
mysqli_stmt_bind_param($statement, "s", $username);

另請注意，您需要檢查插入查詢的返回狀態。 在多用戶系統中，另一個用戶可以在 SELECT 運行和 INSERT 運行之間插入一條記錄。 所以你需要確保你的數據庫在username字段上有UNIQUE約束，然后檢查你的INSERT是否成功。

當您稍后檢查用戶提供的密碼是否正確時，您需要選擇密碼：

$sql = "SELECT pwd FROM usersvet WHERE userName=?";
mysqli_stmt_bind_param($statement, "s", $username);

這將返回散列密碼，然后您可以將其與用戶鍵入的內容進行比較：

if (password_verify($typedPassword, $hashedPasswordFromDatabase)) {
    // correct password provided
} else {
    // wrong password provided
}