警告: mysqli_stmt_bind_param()：类型定义字符串中的元素数与中的绑定变量数不匹配
警告: mysqli_stmt_bind_param()：类型定义字符串中的元素数与中的绑定变量数不匹配
提问于 2019-01-18 03:04:22
我正在用Php和MySqli做一个登录系统，我只是完成了注册功能。
在设置了error_reporting(E_ALL)之后，我得到了前面提到的错误，我得到的两个错误中的一个是，即使我的if语句在被发送到数据库之前没有通过输入。这是我的代码：
<?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($dbConnection);
}前面提到的error-array被用来以div的形式向页面发送视觉反馈，代码如下：
<?php if (!empty($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 ?>回答 1
Stack Overflow用户
回答已采纳
发布于 2019-01-18 03:38:52
如果您的目标是确定用户名是否已经存在，则不必关心密码。
你得到了这个：
$sql = "SELECT userName FROM usersvet WHERE userName=? AND pwd=?";
mysqli_stmt_bind_param($statement, "s", $username);你想要这个：
$sql = "SELECT userName FROM usersvet WHERE userName=?";
mysqli_stmt_bind_param($statement, "s", $username);还要注意，您需要检查insert查询的返回状态。在多用户系统中，另一个用户可以在SELECT运行和INSERT运行之间插入一条记录。因此，您需要确保您的数据库在username字段上有一个惟一的约束，然后检查您的插入是否成功。
当您稍后检查用户提供的密码是否正确时，您需要选择密码：
$sql = "SELECT pwd FROM usersvet WHERE userName=?";
mysqli_stmt_bind_param($statement, "s", $username);这将返回散列密码，然后您可以将其与用户键入的密码进行比较：
if (password_verify($typedPassword, $hashedPasswordFromDatabase)) {
// correct password provided
} else {
// wrong password provided
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54242636
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