Php 使用公式将用户插入数据库无效

我想检查我添加到数据库中的用户是否已经存在，以及是否在公式中键入了内容。当检查结果正确时，我在公式中输入的数据应该被添加到我的数据库中，但事实并非如此。它与while语句有关，因为当我删除这些行时，它可以工作，但每次我刷新页面时，它都会添加一个用户

    <?php

                    $host = 'localhost';
                    $user = 'root';
                    $password = 'FIoLTAZo1pEar83N';
                    $dbname = 'studenten';

                    $dsn = 'mysql:host='. $host .';dbname='. $dbname;
                    $pdo = new PDO($dsn, $user, $password);
                    $pdo->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_OBJ);
                    $pdo->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);

                    if(isset($_POST["u_name"]) && (null !== $_POST["u_name"])){
                        if ($_SERVER["REQUEST_METHOD"] == "POST") {
                            $user_name = $_POST["u_name"];
                            $first_name = $_POST["vorname"];
                            $last_name = $_POST["nachname"];
                            $user_password = $_POST["passwort"];
                        }
                    }

                    if(isset($_POST["u_name"]) && (null !== $_POST["u_name"])){
                        $suchen = $_POST['u_name'];
                        $sql = 'SELECT * FROM studenten WHERE user_name LIKE ?';
                        $stmt = $pdo->prepare($sql);
                        $stmt->execute([$suchen]);
                        while($row = $stmt->fetch(PDO::FETCH_ASSOC)){
                            if($row['user_name'] == null){
                                $sql = 'INSERT INTO studenten(user_name, first_name, last_name, user_password) VALUES(:user_name, :first_name, :last_name, :user_password)';
                                $stmt = $pdo->prepare($sql);
                                $stmt->execute(['user_name' => $user_name, 'first_name' => $first_name, 'last_name' => $last_name, 'user_password' => $user_password]);
                                unset($_POST["u_name"]);
                        }}}                            
                ?>

    <form id='formular_eins' method="post" action="admin.php">
                        <p class="eigenschaften">User Name:</p>
                        <input type='text' name='u_name' placeholder='User Name' required maxlength="7" minlength="7">
                        <br><br>
                        <p class="eigenschaften">Vorname:</p>
                        <input type='text' name='vorname' placeholder='Vorname' required>
                        <br><br>
                        <p class="eigenschaften">Nachname:<p>
                        <input type='text' name='nachname' placeholder='Nachname' required>
                        <br><br>
                        <p class="eigenschaften">Passwort:</p>
                        <input type='text' name='passwort' placeholder='Passwort' maxlength="15" minlength="6" required>
                        <br><br><br>
                        <button type='submit' name="btn" class="btn_speichern" onclick="leeren()">Speichern</button>
                    </form>

---

首先你可以用？？操作员，这将使您的代码更好：

$host = 'localhost';
$user = 'root';
$password = 'FIoLTAZo1pEar83N';
$dbname = 'studenten';

$dsn = 'mysql:host='. $host .';dbname='. $dbname;
$pdo = new PDO($dsn, $user, $password);
$pdo->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_OBJ);
$pdo->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);

$user_name = $_POST["u_name"] ?? $_POST["u_name"];
$first_name = $_POST["vorname"] ?? $_POST["vorname"];
$last_name = $_POST["nachname"] ?? $_POST["nachname"];
$user_password = $_POST["passwort"] ?? $_POST["passwort"];

if($user_name){
  $suchen = $user_name;
  $sql = 'SELECT * FROM studenten WHERE user_name LIKE ?';
  $stmt = $pdo->prepare($sql);
  $stmt->execute([$suchen]);
  while($row = $stmt->fetch(PDO::FETCH_ASSOC)){
      if(!$row){
          $sql = 'INSERT INTO studenten(user_name, first_name, last_name, user_password) VALUES(:user_name, :first_name, :last_name, :user_password)';
          $stmt = $pdo->prepare($sql);
          $stmt->execute(['user_name' => $user_name, 'first_name' => $first_name, 'last_name' => $last_name, 'user_password' => $user_password]);
          unset($_POST["u_name"]);
        }
    }
}

我不认为，你需要使用fetch方法，使用fetchAll

---

在这之后，使用print_r检查将返回$row=$stmt->fetchAll的内容，并检查将返回$stmt->executestation After INSERT命令的内容。有时使用密码有时使用密码可能这是错误的

永远不要以明文或使用MD5/SHA1存储密码！仅存储使用PHP创建的密码哈希，然后可以使用进行验证。看看这篇文章：了解更多关于感谢的信息，这是我的下一步：谢谢，你说的每句话我都试过了，但我想不出问题出在哪里。好的。尝试将显示错误设置为“显示错误”，1；ini设置“显示启动错误”，1；错误报告全部；之后，发送将返回打印的内容，行=$stmt->fetchAll和$stmt->execute。这将有助于了解发生了什么