PHP login wont work

ristic · June 26, 2013

Hi, i've got a problem which i cant find a way around it, tho i think everything is OK with my php code, so i need an opinion if anyone can spot something...

Anyway, this is my php code:

<?php session_start(); ?>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Logovanje</title>
    </head>
    <body>
        <?php
        if (isset($_POST['prijaviSe']))
        {
            include('otvori_konekciju.php');
            $korisnickoIme = $_POST['korisnickoIme'];
            $lozinka = md5($_POST['lozinka']);
            $rezultat = mysql_query("SELECT * FROM korisnik WHERE korisnicko_ime = '$korisnickoIme' AND lozinka = '$lozinka'");
            include('zatvori_konekciju.php');
            if ($rezultat && mysql_num_rows($rezultat) > 0)
            {
                $red = mysql_fetch_assoc($rezultat);
                $_SESSION['id'] = $red['id'];
                $_SESSION['korisnickoIme'] = $red['korisnicko_ime'];
                header('Location: index.php');
            }
            else
            {
                echo '<span style="color: red;">Korisnicko ime ili lozinka nisu ispravni.</span><hr />';
            }
        }
        ?>
        <form action="logovanje.php" method="POST">
            <table>
                <tr>
                    <td>Korisnicko ime:</td>
                    <td><input type="text" name="korisnickoIme" /></td>
                </tr>
                <tr>
                    <td>Lozinka:</td>
                    <td><input type="password" name="lozinka" /></td>
                </tr>
            </table>
            <hr />
            <input type="submit" value="Prijavi se" name="prijaviSe" />
        </form>
    </body>
</html>

and my table with 2 usernames, admin and test, with their password in next column

as u can see, in my php code, my entered password is hashed and then compared with password in the table...but when i type in my login window : "test" , "test"; it prints me out the error message...that's why i changed admins pass to 'admin' (before it was some random hash code like the password for test) and it's still not working...i really don't know what's the problem, and because of that i cant continue my work....Can anyone help? Thank you.

ginerjm · June 26, 2013

echo some information during the whole process so we can see what is happening. Be sure to echo out the value of the username and the password that you are using in your query.

ristic · June 26, 2013

i wrote for input "test" for username and "test" for password, and the values it echoed are the same like those in the table:

test

098f6bcd4621d373cade4e832627b4f6

i have no idea why the IF part wont go through, seems like that the condition in IF( ) is not TRUE....and it is jumping to ELSE part...

Edited June 26, 2013 by ristic

ginerjm · June 26, 2013

Can I see the code with the echo statements added? Add some extra echos to show how you proceeded through the code.

ristic · June 26, 2013

<?php session_start(); ?>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Logovanje</title>
    </head>
    <body>
        <?php
        if (isset($_POST['prijaviSe']))
        {
            include('otvori_konekciju.php');
            $korisnickoIme = $_POST['korisnickoIme'];
            $lozinka = md5($_POST['lozinka']);
            echo "<p>$korisnickoIme";                           // this does get written
            echo "<p>$lozinka <p>";
            $rezultat = mysql_query("SELECT * FROM korisnik WHERE korisnicko_ime = '$korisnickoIme' AND lozinka = '$lozinka'");
            include('zatvori_konekciju.php');
            if ($rezultat && mysql_num_rows($rezultat) > 0)
            {
                $red = mysql_fetch_assoc($rezultat);
                $_SESSION['id'] = $red['id'];
                echo "This is something random in IF part <p>"; // this does NOT get written
                $_SESSION['korisnickoIme'] = $red['korisnicko_ime'];
                header('Location: index.php');
            }
            else
            {
                echo "ELSE part of the code";                   // written
                echo "<p>$korisnickoIme";
                echo "<p>$lozinka <p>";
                echo '<span style="color: red;">Korisnicko ime ili lozinka nisu ispravni.</span><hr />';
            }
        }
        ?>
        <form action="logovanje.php" method="POST">
            <table>
                <tr>
                    <td>Korisnicko ime:</td>
                    <td><input type="text" name="korisnickoIme" /></td>
                </tr>
                <tr>
                    <td>Lozinka:</td>
                    <td><input type="password" name="lozinka" /></td>
                </tr>
            </table>
            <hr />
            <input type="submit" value="Prijavi se" name="prijaviSe" />
        </form>
    </body>
</html>

output:

test

098f6bcd4621d373cade4e832627b4f6

ELSE part of the code

test

098f6bcd4621d373cade4e832627b4f6

Korisnicko ime ili lozinka nisu ispravni.

ginerjm · June 27, 2013

Do this - replace this line:

echo "ELSE part of the code"; // written

with:

echo "ELSE part of the code - error message is: ".MySQL_error();

Your query is not returning a result OR it completely failed. The 'MySQL_error()' will tell us which one it is.

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PHP login wont work

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