登录表单错误。调用未定义的函数mysql_query()

时间:2017-09-06 03:32:26

标签: php mysql xampp

美好的一天。我试图使用php和xampp来创建一个登录表单来运行代码。我已经在phpMyAdmin中有一个数据库,我收到此错误

连接!

致命错误:未捕获错误:在C:\ xampp \ htdocs \ login \ connection.php中调用未定义函数mysql_query():34堆栈跟踪:在C:\ xampp \ htdocs \ login \中抛出#0 {main}第34行的connection.php

每次执行此代码

<?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");

}


?>

我尝试在youtube上搜索如何修复错误,但他们只是告诉我的php版本已过时/更新。我很困惑。

1 个答案:

答案 0 :(得分:0)

如果您已设置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");

    }