sql injection login form error

Question

I am working on tutorial about how to make an sql injection with old php code. I have this code that I am testing on it, but I have an error that said:

Fatal error: Call to a member function fetch_assoc() on a non-object

This is the old code:

<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="root"; // Mysql password 
$db_name="graphic_db"; // Database name 
$tbl_name="login"; // Table name 
//if(!session_is_registered(myusername)){
//header("location:index.html");
  $con=mysqli_connect($host,$username,$password,$db_name);



if(isset($_POST['login_submit'])){

    if($_POST['username'] != '' && $_POST['password']!=''){
        if(!isset($_SESSION)) 
    { 
        session_start();
        //session_register('username'); 
    } 

        $result = mysqli_query($con, "SELECT * FROM login WHERE username='" . $_POST["username"] . "' and password = '". $_POST["password"]."'");
        while($row  = $result->fetch_assoc($result) ){
        if(is_array($row)) {
            $_SESSION["username"] = $row[$_POST["username"]];
            $_SESSION['username'] = $_POST["username"];

            header("Location:home.php");
            } 
            else {
            $message = "Invalid Username or Password!";

            }

        }
        }

        }

?>

Answer 1

You have mixing Object oriented style and Procedural style in your code .Use only one style as

In Object oriented style

$result = $mysqli->query( "SELECT * FROM login WHERE username='" . $_POST["username"] . "' and password = '". $_POST["password"]."'");
 while($row  = $result->fetch_assoc() ){

In Procedural style

 $result = mysqli_query($con, "SELECT * FROM login WHERE username='" . $_POST["username"] . "' and password = '". $_POST["password"]."'");
 while($row  = mysqli_fetch_assoc($result) ){

Read http://php.net/manual/en/mysqli-result.fetch-assoc.php

Answer 2

试试这段代码$row=mysqli_fetch_array($result,MYSQLI_ASSOC);

Answer 3

    $host="localhost"; // Host name 
    $username="root"; // Mysql username 
    $password="root"; // Mysql password 
    $db_name="graphic_db"; // Database name 
    $tbl_name="login"; // Table name 
    session_start();
    $mysqli = new mysqli($host, $username, $password, $db_name);

    if(isset($_POST['login_submit'])){
        $stmt = $mysqli->prepare("SELECT * FROM login WHERE email=? AND  password=? ");
        $email = $_POST['username'];
        $password = md5($_POST['password']);
        $stmt->bind_param('ss', $email, $password);
        $stmt->execute();
        $stmt->bind_result($email, $password);
        $stmt->store_result();

        if($stmt->num_rows == 1)  //To check if the row exists
        {
            while($stmt->fetch()) //fetching the contents of the row
            {
                $_SESSION['Logged'] = 1;
                   $_SESSION['Email'] = $email;
                   header('Location: home.php');
                   exit();
            }

        }
        else 
        {
            echo "Wrong Username or Password!";
        }
        $stmt->close();
        $stmt->free_result();
    }

Use prepared query for your sql injection code. it is best practice for it.

Prepared statements are very useful against SQL injections, because parameter values, which are transmitted later using a different protocol, need not be correctly escaped. If the original statement template is not derived from external input, SQL injection cannot occur.