Php 检查用户名是否存在或不工作？

我已经检查了其他答案，并尝试将其复制到我的答案中，但我得到了一个未识别的

行

错误

<?php
session_start();

if(isset($_SESSION['user_id'])) {
    header("Location: index.php");
}

include_once 'dbconnect.php';

//set validation error flag as false
$error = false;

//check if form is submitted
if (isset($_POST['signup'])) {
    $first_name = mysqli_real_escape_string($con, $_POST['first_name']);
    $last_name = mysqli_real_escape_string($con, $_POST['last_name']);
    $username = mysqli_real_escape_string($con, $_POST['username']);
    $email = mysqli_real_escape_string($con, $_POST['email']);
    $password = mysqli_real_escape_string($con, $_POST['password']);
    $cpassword = mysqli_real_escape_string($con, $_POST['cpassword']);
    $quote = mysqli_real_escape_string($con, $_POST['quote']);
    $who = mysqli_real_escape_string($con, $_POST['who']);

    //name can contain only alpha characters and space
    if ($_POST['username'] == $row['username']){
        $error = true;
        $username_error = '<div class="error-notice">
                            <div class="oaerror danger">
                            <strong>Uh oh!</strong> - Username is already in use.
                            </div>';

    }

---

没关系，我明白了。我必须在代码上方执行此操作

$result = mysqli_query($con, "SELECT * FROM users WHERE username = '" . $username. "'");
    if($row = mysqli_fetch_array($result)){

你把$row分配到哪里？哦，我没有。。。我应该将其分配给什么？您应该使用而不是连接查询。即使是
mysqli\u real\u escape\u string（）
对于注入攻击也不是100%安全的。我还不知道如何做到这一切。我是个新手，所以我的计划是一步一步地进步。