使用PHP和MySQL检查数据库中是否存在用户名

已经为此挣扎了几天，读到了大量类似的问题，但仍然无法解决。我试图通过html表单获取用户信息并用它更新我的数据库，但前提是用户名不存在。。。 问题1：不管怎样，它都可以用相同的用户名创建条目 问题2：我一直收到错误警告：mysqli_num_rows（）期望参数1是mysqli_result，bool给定

我确信他们是有联系的，但就我个人而言，我不知道解决办法是什么

<?php
//declare variables
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "mypassword";
$db = "user_info";

if (!empty($_POST)) {

  //connect to mysqli
  $mysqli = new mysqli($dbhost, $dbuser, $dbpass, $db);

  //check connection
  if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_errno . ': ' . $mysqli->connect_error );
  }

  //declare user variables
  $first_name = $_POST['first_name'];
  $last_name = $_POST['last_name'];
  $user_name = $_POST['user_name'];
  $pass = $_POST['pass'];
  $address = $_POST['address'];
  $phone = $_POST['phone'];

  //get data from form
  $sql = "INSERT INTO users (first_name, last_name, user_name, pass, address, phone) VALUES
    ('{$mysqli->real_escape_string($first_name)}',
    '{$mysqli->real_escape_string($last_name)}',
    '{$mysqli->real_escape_string($user_name)}',
    '{$mysqli->real_escape_string($pass)}',
    '{$mysqli->real_escape_string($address)}',
    '{$mysqli->real_escape_string($phone)}')";

// ** HERE IS WHERE I AM HAVING ISSUES ** 

  //query to see if entered username already exists
  $result = $mysqli->query("SELECT * FROM users WHERE user_name =$user_name");

  //alert if user name taken
  if (mysqli_num_rows($result) > 0) {
    echo "<b>Username already taken! Please select another.</b>";
  } else {

  //continue to insert into database if username is unique
  $insert = $mysqli->query($sql);

  //print response from mysql
  if ($insert) {
    echo "Success! Row ID: {$mysqli->insert_id}";
  } else {
    die("error: {$mysqli->errno}");
  }
}
  //close our app
  $mysqli->close();

}
?>

像这样试试

<?php


 if (!empty($_POST)) {

    //connect to mysqli
    $mysqli = new mysqli($dbhost, $dbuser, $dbpass, $db);

    //check connection
    if ($mysqli->connect_error) {
        die('Connect Error: ' . $mysqli->connect_errno . ': ' . $mysqli->connect_error );
    }

    //declare user variables
    $first_name = $_POST['first_name'];
    $last_name = $_POST['last_name'];
    $user_name = $_POST['user_name'];
    $pass = $_POST['pass'];
    $address = $_POST['address'];
    $phone = $_POST['phone'];

    $query = $mysqli->prepare("SELECT count(*) FROM users WHERE user_name = ?");
    $query->bind_param('s', $user_name);
    $query->bind_result($cnt);
    $query->execute();
    $query->store_result();
    $query->fetch();
    $query->close();

    if ($cnt > 0) {
        echo "<b>Username already taken! Please select another.</b>";
    } else {
        $query = $mysqli->prepare("INSERT INTO users (first_name, last_name, user_name, pass, address, phone) VALUES (?,?,?,?,?,?)");
        $query->bind_param('ssssss', $first_name, $last_name, $user_name, $pass, $address, $phone);
        $query->execute();
        if ($query->execute()) {
            echo "Success! Row ID: {$query->insert_id}";
        } else {
            die("error: {$query->errno}");
        }
        $query->close();
    }
}

连接POST变量不是一个好主意，使用prepared语句我不理解SELECT的意义。如果插入失败（因为您已将列指定为唯一的），那么您就知道需要知道的一切。您的问题称为SQL注入。之所以会看到这个奇怪的消息，是因为您使用SQL注入破坏了自己的SQL，并且没有启用错误报告，所以mysqli不会告诉您错误。这也是一个XY问题，因为这个选择不应该存在。谢谢大家的帮助。一般来说，我对PHP和编程都非常陌生，所以我将接受你们所说的，并在我的项目中工作！请不要回答封闭式问题，当您回答时，请至少解释您的解决方案。