php - 登录表单错误。调用未定义的函数 mysql_query()

Question

再会。我正在尝试使用 php 和 xampp 制作登录表单来运行代码。我已经在 phpMyAdmin 中有一个数据库，我收到了这个错误

连接的!

fatal error : Uncaught Error :调用 C:\xampp\htdocs\login\connection.php:34 中未定义的函数 mysql_query() 堆栈跟踪:#0 {main} 在 C:\xampp\htdocs\login\connection.php 中抛出在第 34 行

每次我执行这段代码

<?php
$host = 'localhost';
$uname = 'root';
$pword = '';
$db = 'login';
$port = '3306';

$link = mysqli_init();
$success = mysqli_real_connect(
       $link,
       $host,
       $uname,
       $pword,
       $db,
       $port
    );


if(isset($_POST['user']))
{
    $username = $_POST['user'];
}
if(isset($_POST['pass']))
{
    $password = $_POST['pass'];
}



$con = new mysqli($host, $uname, $pword, $db) or die("Connection failed". mysqli_error());

echo ("Connected!");

$result = mysql_query($success. "select * from db_login where username = $username and password = $password");

$row = mysql_fetch_array($result);

if($row['username'] == $username && $row['password'] == $password )
{
    echo("Welcome!". $row['username']);

}
else
{
    echo("Login failed");

}


?>

我试图在 youtube 上搜索如何修复错误，但他们只是告诉我我的 php 版本已过时/更新。我糊涂了。

Answer 1

如果您设置了与数据库的正确连接....

$host = 'localhost';
$uname = 'root';
$pword = '';
$db = 'login';
$success = mysqli_connect($host,$uname,$pword,$db);
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


if(isset($_POST['user']))
{
    $Username = $_POST['user'];
}
if(isset($_POST['pass']))
{
    $Password = $_POST['pass'];
}
// removes backslashes
 $username = stripslashes($Username);
 //escapes special characters in a string
 $username = mysqli_real_escape_string($success,$username);
 $password = stripslashes($Password);
 $password = mysqli_real_escape_string($success,$password);

$result = mysqli_query($success. "select * from db_login where username = $username and password = $password") or die(mysql_error());
$row = mysqli_num_rows($result);

    if($rows==1){
        echo("Welcome!". $row['username']);

    }
    else
    {
        echo("Login failed");

    }