Simple PHP/MySQL Login script - always failing

Question

I have been recently making a project - an online quiz, so I decided to make a registration/login system as well. I have faced some issues with the login part, so I decided to do everything from scratch, but the problem is still there.

The registration works fine, I managed to register 10 users without problems, but when I try to login it is always showing - Wrong login or password! though everything is correctly input and there are no duplicate records in my DB.

Here is the connection to the database:

<?php
$hostname = "localhost";
$username = "user";
$pass = "pass";
$db = "testdb";

$connect = mysqli_connect($hostname, $username, $pass) or die("Something went wrong!");

mysqli_select_db($connect, $db) or die("Couldn`t connect to database!");
?>`

login.php

<html>
<head>
<?php
require ('db.php');

if(isset($_POST['login']))
{
    $login = mysqli_real_escape_string($connect, $_POST['login']);
    $password = mysqli_real_escape_string($connect, $_POST['password']);

    if($login == NULL || $password == NULL)
    {
        echo "All fields must be completed!";
        exit();
    }

    else
    {
        $sql = "SELECT * FROM `users` WHERE `login` = '".$login."' AND `password` = '".$password."'";
        $result = mysqli_query($connect, $sql);

        if($sql->num_rows > 0)
        {
            echo "Successfully logged in!";
        }

        else
        {
            echo "Wrong login or password!";
        }
    }
}
?>
</head>

<body>
<div align="center">
<h1>Login</h1>
</div>

<form action="login.php" method="post" align="center">
<input name="login" type="text" placeholder="Login"><br><br>
<input name="password" type="password" placeholder="Password"><br><br>

<input name="login" type="submit" value="Login">
</form>

</body>
</html>

I am so desperate, I have been searching for a solution for days! I really do not know why is this happening. I would appreciate any help!

Alan Tan · Accepted Answer · 2015-12-23 09:34:06Z

2

Your submit button has the same name as your username field.

<input name="login" type="submit" value="Login">

change it to something else.

Eg:

<input name="loginBtn" type="submit" value="Login">

And also change

if($sql->num_rows > 0)

to

if($result->num_rows > 0)

answered Dec 23, 2015 at 9:34

Alan Tan

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3 Comments

Max775 Over a year ago

It works!!! Thank you very very much! Omg! I will NEVER give same name to buttons and fields :)

Alan Tan Over a year ago

@Max775 No probs, do make sure to vote for the best answer and all the best to you :)

Alan Tan Over a year ago

@DanyalSandeelo Hey, sorry I didn't give you credit for noticing the flaw in $sql->num_rows, but your answer was count($result) if I remembered correctly. Before you edited your answer :)

Danyal Sandeelo · Accepted Answer · 2015-12-23 09:32:21Z

2

$sql is string, you need to perform the check on resultSet

Change

 if($sql->num_rows > 0)

to

 if($result->num_rows > 0)

answered Dec 23, 2015 at 9:32

Danyal Sandeelo

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