久久久久久久av_日韩在线中文_看一级毛片视频_日本精品二区_成人深夜福利视频_武道仙尊动漫在线观看

mysqli_stmt_close() 期望參數 1 是 mysqli_stmt，布爾值

2023-07-29 php問題 html5模板網

mysqli_stmt_close() expects parameter 1 to be mysqli_stmt, boolean(mysqli_stmt_close() 期望參數 1 是 mysqli_stmt，布爾值)

本文介紹了mysqli_stmt_close() 期望參數 1 是 mysqli_stmt，布爾值的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！

問題描述

這在大約 5 分鐘前還在工作，但突然停止了.這是一個簡單的登錄表單，請看下面的代碼

<!DOCTYPE html><html lang="zh-cn"><頭><meta charset="UTF-8"><title>登錄</title><link rel="stylesheet" ><style type="text/css">正文{字體:14px sans-serif;}.wrapper{ 寬度:350px;填充:20px；}</風格><身體><div class="wrapper"><h2>登錄</h2><p>請填寫您的登錄憑據.</p><form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>"方法=發布"><div class="form-group <?php echo (!empty($username_err)) ? 'has-error' : ''; ?>"><label>用戶名</label><input type="text" name="username" class="form-control" value="<?php echo $username; ?>"><span class="help-block"><?php echo $username_err;?></span>

<div class="form-group <?php echo (!empty($password_err)) ? 'has-error' : ''; ?>"><label>密碼</label><input type="password" name="password" class="form-control"><span class="help-block"><?php echo $password_err;?></span>

<div class="form-group"><input type="submit" class="btn btn-primary" value="登錄">

<p>還沒有帳戶?<a href="register.php">立即注冊</a>.</p></表單>

我得到的錯誤是 Warning: mysqli_stmt_close() 期望參數 1 是 mysqli_stmt，布爾值在第 83 行的/opt/lampp/htdocs/magic/client/login.php 中給出

這在大約 5 分鐘前還在工作，但突然停止了，不知道為什么.有什么想法嗎?

當它工作時，它只是將用戶重定向到welcome.php，現在它不是，并給我那個錯誤.代碼取自 https://www.tutorialrepublic.com/php-tutorial/php-mysql-login-system.php

解決方案

看看你在問什么:

if($stmt = mysqli_prepare($link, $sql)){//[...]}mysqli_stmt_close($stmt);

您總是嘗試關閉您的語句，即使它沒有成功創建.您需要移動該關閉嘗試:

if($stmt = mysqli_prepare($link, $sql)){//[...]mysqli_stmt_close($stmt);}

此外，您首先收到此錯誤的事實意味著您的 SQL 查詢未能準備好，因此其中存在某種錯誤.您目前沒有檢查這一點，如果您想找出為什么您的代碼不起作用，這可能會讓您感到困難.我建議為此添加一個檢查:

if($stmt = mysqli_prepare($link, $sql)){//[...]mysqli_stmt_close($stmt);} 別的 {回聲查詢有問題:".mysqli_error($link);}

請注意，在生產代碼中，您不想直接輸出 mysqli_error() ，因為它可能會顯示有關您的數據庫的信息，而這些信息不是任何人的業務而是您的業務，因此請僅在此期間執行此操作你正在調試.

this was working about 5 minutes ago and has suddently stopped. It is a simple login form, please see code below

<?php
// Initialize the session
session_start();

// Check if the user is already logged in, if yes then redirect him to welcome page
if(isset($_SESSION["loggedin"]) && $_SESSION["loggedin"] === true){
    header("location: welcome.php");
    exit;
}

// Include config file
require_once "db/config.php";

// Define variables and initialize with empty values
$username = $password = "";
$username_err = $password_err = "";

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){

    // Check if username is empty
    if(empty(trim($_POST["username"]))){
        $username_err = "Please enter username.";
    } else{
        $username = trim($_POST["username"]);
    }

    // Check if password is empty
    if(empty(trim($_POST["password"]))){
        $password_err = "Please enter your password.";
    } else{
        $password = trim($_POST["password"]);
    }

    // Validate credentials
    if(empty($username_err) && empty($password_err)){
        // Prepare a select statement
        $sql = "SELECT id, username, password FROM users WHERE username = ?";

        if($stmt = mysqli_prepare($link, $sql)){
            // Bind variables to the prepared statement as parameters
            mysqli_stmt_bind_param($stmt, "s", $param_username);

            // Set parameters
            $param_username = $username;

            // Attempt to execute the prepared statement
            if(mysqli_stmt_execute($stmt)){
                // Store result
                mysqli_stmt_store_result($stmt);

                // Check if username exists, if yes then verify password
                if(mysqli_stmt_num_rows($stmt) == 1){                    
                    // Bind result variables
                    mysqli_stmt_bind_result($stmt, $id, $username, $hashed_password);
                    if(mysqli_stmt_fetch($stmt)){
                        if(password_verify($password, $hashed_password)){
                            // Password is correct, so start a new session
                            session_start();

                            // Store data in session variables
                            $_SESSION["loggedin"] = true;
                            $_SESSION["id"] = $id;
                            $_SESSION["username"] = $username;                            

                            // Redirect user to welcome page
                            header("location: welcome.php");
                        } else{
                            // Display an error message if password is not valid
                            $password_err = "The password you entered was not valid.";
                        }
                    }
                } else{
                    // Display an error message if username doesn't exist
                    $username_err = "No account found with that username.";
                }
            } else{
                echo "Oops! Something went wrong. Please try again later.";
            }
        }

        // Close statement
        mysqli_stmt_close($stmt);
    }

    // Close connection
    mysqli_close($link);
}
?>

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Login</title>
    <link rel="stylesheet" >
    <style type="text/css">
        body{ font: 14px sans-serif; }
        .wrapper{ width: 350px; padding: 20px; }
    </style>
</head>
<body>
    <div class="wrapper">
        <h2>Login</h2>
        <p>Please fill in your credentials to login.</p>
        <form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
            <div class="form-group <?php echo (!empty($username_err)) ? 'has-error' : ''; ?>">
                <label>Username</label>
                <input type="text" name="username" class="form-control" value="<?php echo $username; ?>">
                <span class="help-block"><?php echo $username_err; ?></span>
            </div>    
            <div class="form-group <?php echo (!empty($password_err)) ? 'has-error' : ''; ?>">
                <label>Password</label>
                <input type="password" name="password" class="form-control">
                <span class="help-block"><?php echo $password_err; ?></span>
            </div>
            <div class="form-group">
                <input type="submit" class="btn btn-primary" value="Login">
            </div>
            <p>Don't have an account? <a href="register.php">Sign up now</a>.</p>
        </form>
    </div>    
</body>
</html>

The error I get is Warning: mysqli_stmt_close() expects parameter 1 to be mysqli_stmt, boolean given in /opt/lampp/htdocs/magic/client/login.php on line 83

This was working about 5 minutes ago and has suddenly stopped and not sure why. Any ideas?

When it worked, it simply redirected the user to welcome.php and now it is not, and gives me that error. The code was taken of https://www.tutorialrepublic.com/php-tutorial/php-mysql-login-system.php

解決方案

Look at what you're asking:

if($stmt = mysqli_prepare($link, $sql)){
    // [...]
}

mysqli_stmt_close($stmt);

You always try to close your statement, even if it wasn't successfully created. You need to move that close attempt:

if($stmt = mysqli_prepare($link, $sql)){
    // [...]

    mysqli_stmt_close($stmt);
}

Additionally, the fact that you got this error in the first place means your SQL query failed to be prepared, so there's some kind of error in there. You are currently not checking for that, which can make life difficult for you if you want to figure out why your code isn't working. I recommend adding a check for that:

if($stmt = mysqli_prepare($link, $sql)){
    // [...]

    mysqli_stmt_close($stmt);
} else {
    echo "Something's wrong with the query: " . mysqli_error($link);
}

Note that in production code, you don't want to output the mysqli_error() directly since it may reveal information about your database that isn't anybody's business but yours, so only do this while you're debugging.

這篇關于mysqli_stmt_close() 期望參數 1 是 mysqli_stmt，布爾值的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！

【網站聲明】本站部分內容來源于互聯網,旨在幫助大家更快的解決問題，如果有圖片或者內容侵犯了您的權益，請聯系我們刪除處理，感謝您的支持！

上一篇：CodeIgniter Active Record:一次加載一行下一篇：通過 SSH 通過 PHP 連接到 mysql 數據庫

相關文檔推薦

mysql 中的 store_result() 和 get_result() 返回 false

store_result() and get_result() in mysql returns false(mysql 中的 store_result() 和 get_result() 返回 false)

調用未定義的函數 mysqli_result::num_rows()

Call to undefined function mysqli_result::num_rows()(調用未定義的函數 mysqli_result::num_rows())

PHP 準備好的語句問題

PHP Prepared Statement Problems(PHP 準備好的語句問題)

mysqli_fetch_array 只返回一個結果

mysqli_fetch_array returning only one result(mysqli_fetch_array 只返回一個結果)

PHP MySQLi 多次插入

PHP MySQLi Multiple Inserts(PHP MySQLi 多次插入)

如何確保 MySQL 中的值在 PHP 中保持其類型?

How do I make sure that values from MySQL keep their type in PHP?(如何確保 MySQL 中的值在 PHP 中保持其類型?)

欄目導航

前端問題解決 Java問題 php問題 Python問題 C#/.NET問題 C/C++問題移動開發問題數據庫問題