[制作CRUD应用程序时与php中的mysqli有关的错误[重复]

问题描述 投票:0回答:1

所以我是PHP的初学者,并且正在使用MySQL制作一个简单的CRUD应用程序。制作CRUD应用程序表已完成,并且已成功将其与我的MySQL数据库连接。但是,当我在服务器上加载页面时,遇到了无法调试的错误。这是我遇到错误的CRUD表的代码的一部分(代码的57-74行):

// Attempt select query execution my
                    $sql = "SELECT * FROM students";
                    if($result = $mysqli->query($sql)){
                        if($result->num_rows > 0){
                            echo "<table class='table table-bordered table-striped'>";
                                echo "<thead>";
                                    echo "<tr>";
                                        echo "<th>id</th>";
                                        echo "<th>Name</th>";
                                        echo "<th>Father's Name</th>";
                                        echo "<th>Mother's Name</th>";
                                        echo "<th>Email</th>";
                                        echo "<th>Contact</th>";
                                        echo "<th>Alt. Contact</th>";
                                        echo "<th>Alt. Contact</th>";
                                    echo "</tr>";
                                echo "</thead>";
                                echo "<tbody>";

这是我得到的错误:

Notice: Undefined variable: mysqli in /Applications/XAMPP/xamppfiles/htdocs/student/options2.php on line 59

Fatal error: Uncaught Error: Call to a member function query() on null in /Applications/XAMPP/xamppfiles/htdocs/student/options2.php:59 Stack trace: #0 {main} thrown in /Applications/XAMPP/xamppfiles/htdocs/student/options2.php on line 59
php mysql mysqli crud
1个回答
0
投票

您丢失了PHP和MySQL之间数据库连接的最重要的一行。 mysqli类,表示PHP和MySQL数据库之间的连接。另外,启用mysqli错误报告,以便您可以轻松找出错误。查看更多:https://www.php.net/manual/en/mysqli.query.php

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("your_host", "your_user", "your_password", "your_database");

// Change character set to utf8
$mysqli->set_charset("utf8mb4");
© www.soinside.com 2019 - 2024. All rights reserved.