不将数据输入数据库的PHP代码

Question

我无法将表格发送到数据库。我已经试了一个小时来解决这个问题；我用“echo mysqli_error（）；”将代码打了个谜，但并没有收到任何错误，只是数据库中缺少新数据（似乎提交表单也并没有重新加载页面来发送信息）。数据库使用config.php、connectDB.php和header.php为我的其余页面工作，因此问题不在数据库设置中。我相信错误在“isset（$\u POST['submit']）”中，因为它不会用header（）重新加载页面，但我提供了我的其余代码以防万一。（我知道这段代码不受SQL注入的保护，表单验证是我修复后的下一步）：

<?php
require_once ("Includes/config.php"); 
require_once  ("Includes/connectDB.php");
include("Includes/header.php");
if (isset($_POST['submit'])){
                $name = $_POST['name'];
                $email = $_POST['email'];
                $content = $_POST['content'];
                $query = "INSERT INTO requests (name, email, content) VALUES (?, ?, ?)";

                $statement = $databaseConnection->prepare($query);
                $statement->bind_param('sss', $name, $email, $content);
                header('Location: /index.php');

                $statement->execute()
                $statement->store_result();
                setcookie("nameErr"," ", time()+3600);
                setcookie("emailErr"," ", time()+3600);
                setcookie("contentErr"," ", time()+3600);
                setcookie("contentSucc","Announcement Request Successful", time()+3600);
                header('Location: /request.php');

}
?>
<div id="main">
        <ol>
            <li>
                <label for="name">Name:</label> 
                <input type="text" name="name" value="" id="name" />
                <span class="error">* <?php if (isset($_COOKIE["nameErr"])){echo $_COOKIE["nameErr"];}?></span>
            </li>
            <li>
                <label for="email">Email:</label>
                <input type="text" name="email" value="" id="email" style = "position: relative; left: 3px;"/>
                <span class="error" style = "position: relative; left: 3px;">* <?php if (isset($_COOKIE["emailErr"])){echo $_COOKIE["emailErr"];}?></span>
            </li>
            <li>
                    <label for="content">Requested Announcement:</label><br>
                    <textarea rows="18" cols="140" name="content" id="content"></textarea>
                    <br>
                    <span class="error"> <?php if (isset($_COOKIE["contentErr"])){echo $_COOKIE["contentErr"];}?></span>
                    <span class="error"> <?php if (isset($_COOKIE["contentSucc"])){echo $_COOKIE["contentSucc"];}?></span>
            </li>
        </ol>
        <input type="submit" name="submit" value="Submit " style = "position: relative; left: 40px;" /> 
        <!-- onclick="window.location='request.php';" -->   
        <p>
            <a class="cancel" href="index.php" style = "position: relative; left: 40px;">Cancel</a>
        </p>
</div>

Answer 1

1. 在执行SQL之前，您将重定向到另一个网页。

改变这个

            $statement->bind_param('sss', $name, $email, $content);
            header('Location: /index.php');

            $statement->execute()

对此

            $statement->bind_param('sss', $name, $email, $content);
            $statement->execute();
            $statement->close();

            header('Location: /index.php');