php - 我在插入“对未定义方法mysqli_stmt::get_result()的调用”上不断收到此错误

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Fatal error: Call to undefined method mysqli_stmt::get_result()
                                
                                    （2个答案）
                                
                        
                                6年前关闭。
            
                    
我不确定为什么会一直收到此错误，但是我需要一些帮助...我只是在尝试允许允许插入之前，检查数据库以查看记录是否存在。

$input_errors = array();

if (!empty($_POST['username'])) {
    $user = $_POST['username'];
} else {
    $input_errors['username'] = "Must fill out username";
}

$email = filter_input(INPUT_POST, 'usermail', FILTER_VALIDATE_EMAIL);
if (false === $email) {
    $input_errors['usermail'] = "Not a valid email address";
}

if(count($input_errors) > 0) {
    print_r($input_errors); die();
}

$sql = "SELECT COUNT(*) as amount FROM people WHERE username = ?
       OR email = ?";
if ($stmt = $mysqli->prepare($sql)) {

    $stmt->bind_param("ss", $user, $email);
    $stmt->execute();
    $results = $stmt->get_result();
    $data = mysqli_fetch_assoc($results);

    if ($data['amount'] > 0)
    {
        print "User already exists";
    }
}

else {
$stmt = $mysqli->stmt_init();
if (!$stmt) {
    echo "Init failed";
} else {
    $cmd = "INSERT INTO people (username, email, sign_up_date) VALUES (?, ?, NOW() )";
    if ($stmt->prepare($cmd)) {
        $stmt->bind_param('ss', $user, $email );
        $stmt->execute();

        echo $stmt->affected_rows . " row(s) inserted";

        $stmt->close();

    } else {
        echo "Prepare failed";
    }
    mysqli_close($mysqli);
    }
}

如果有人可以伸出援手，那就太好了。

最佳答案

由于某些原因，我无法忍受这么长而多风的代码。
您实际需要的只是

$cmd = "INSERT IGNORE INTO people (username, email, sign_up_date) VALUES (?, ?, NOW() )";
$stmt->prepare($cmd);
$stmt->bind_param('ss', $user, $email);
$stmt->execute();
if (!$stmt->affected_rows)
{
    print "User already exists";
}

无需麻烦的额外选择查询。
无需麻烦的绑定结果。
无需麻烦处理大量嵌套的if语句（其逻辑有缺陷，实际上破坏了整个混乱）。