OOP INSERT INTO 语句不起作用

OOP INSERT INTO statement not working

我正在创建一个非常简单的留言板来学习 OOP 风格的编程。目的是显示保存在数据库中的post,在显示的post下面,你可以添加一个新的post到数据库,它也会显示在页面上。

我在调用 index.php 页面上的函数时遇到问题,所以当我单击提交时,它会将文本区域中的内容添加到数据库中。有什么建议吗?

Index.php

<?php

ini_set("display_errors", 1);
error_reporting(E_ALL);

require_once("prikbord.class.php");

$p = new Prikbord();
$bericht = $p->getbericht();

if(isset($_POST['submit'])){
    $p->setbericht($_POST[$bericht]);
}
?>

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
        <title>Prikbord</title>
    </head>

    <body>
        <?php 
            echo $p->showdata();
        ?>

        <form action="" type="post">
            <textarea name="bericht" placeholder="Uw bericht" style="display: block; width: 350px; height: 150px;"></textarea>
            <input type="submit" name="submit" value="Submit">
        </form>
    </body>
</html>

Prikbord.class.php

<?php

class Prikbord {

    private $db;
    private $data = array();

    public function __construct() {
        $this->db = new mysqli("localhost", "root", "root", "prikbord");

        if($this->db->connect_error) {
            die("Fout met database verbinding.");
        }
    }

    public function getbericht() {
        $sql = "SELECT naam, datum, bericht FROM prikbord";

        $result = $this->db->query($sql);

        while($row = $result->fetch_assoc()) {
            $this->data[] = $row;
        }

        return $this->data;
    }

    public function showdata() {
        $return = "";

        foreach($this->data as $item) {
            $return .= "<div>" . $item["naam"] . "</div>";
            $return .= "<div>" . $item["datum"] . "</div>";
            $return .= "<div>" . $item["bericht"] . "</div>";
            $return .= "<hr style='width: 100px; float: left; display: inline;'>";
            $return .= "<br><br>";
        }

        return $return;
    }

    public function setbericht($bericht) {
        $sql = "INSERT INTO prikbord(bericht) VALUES ('" . $bericht . "')";

        $result = $this->db->query($sql);

        if($this->result) {
            echo "Data insert succesfully.";
        }
    }
}

?>

您根本没有发布您的数据。你应该像这样打开你的表格:

<form name="myForm" method="POST">

注:method,不是type