检查PHP中是否有用户名

I am trying to perform a 'Duplicate Username' check using the code below. It continues to tell me that the username is available when in fact it isn't. Can someone tell me what I am doing wrong.

我正在尝试使用下面的代码执行“重复用户名”检查。它继续告诉我,用户名是可用的,实际上它不是。有人能告诉我我做错了什么。

FORM

<form method="post" action="scripts/checkusername.php">                        
       <div class="form-group" hidden>
          <label for="store">User ID</label>
          <input type="text" class="form-control" name="uid" id="uid" value="<?php echo " " . $_SESSION["UserID"] . " "; ?>">
        </div>
        <div class="form-group">
          <label class="control-label">Store URL</label>
             <p class="form-control-static"><b><?php echo isset($row['Username']) ? $row['Username'] : NULL; ?></b>.example.com</p>
        </div>
        <div class="form-group">
             <label for="store">Store Name</label>
             <input type="text" class="form-control" name="username" id="username" value="<?php echo isset($row['Username']) ? $row['Username'] : NULL; ?>">
        </div>
        <div class="form-group">
             <label for="exampleInputEmail1">Email Address</label>
             <input type="email" class="form-control" name="email" id="email" value="<?php echo isset($row['EmailAddress']) ? $row['EmailAddress'] : NULL; ?>">
        </div>

        <button type="submit" class="btn btn-prime">Update Details</button>
</form>

SCRIPT

<?php

    $servername = "localhost";
    $username = "root";
    $password = "root";
    $dbname = "users";

    $conn = new mysqli($servername, $username, $password, $dbname);

    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    if (isset($_POST['username'])) 
    {
        $username = mysql_real_escape_string($_POST['username']);

        if (!empty($username)) 
    {
            $username_query = mysql_query("SELECT *
                                           FROM users
                                           WHERE Username = '$username'");
             $count=mysql_num_rows( $username_query);
             if($count==0)
             {
               echo "Username doesn't exist";
               exit;
             }
            else
            {
              echo "Username already exists";
              exit;
            }
    }
}

?>

1 个解决方案

#1

Since you used new mysqli to open the database connection, you have to use mysqli functions to perform queries, not mysql functions.

由于您使用新的mysqli来打开数据库连接,您必须使用mysqli函数来执行查询,而不是mysql函数。

if (isset($_POST['username'])) 
{
    $username = $_POST['username'];

    if (!empty($username)) 
    {
        $username_query = $conn->prepare("SELECT *
                                       FROM users
                                       WHERE Username = ?");
        $username_query->bind_param("s", $username);
        $username_query->execute();
        $count = $username_query->num_rows;
        if($count==0)
        {
            echo "Username doesn't exist";
            exit;
        }
        else
        {
            echo "Username already exists";
            exit;
        }
    }
}