Help With A Simple Login PHP

MarPlo 1 Junior Poster in Training · 2012-06-21T15:48:48+00:00

Hi,
Change in your form with this:

Username: <input type="text" name="user"/></br>
Password: <input type="text" name="password"/></br>

"name" value and $_POST['value'] must be the same value.

tarekwiz 0 Newbie Poster · 2012-06-21T16:03:08+00:00

thanks but now when i try to save cokkie it says
Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\login.php:5) in C:\xampp\htdocs\login.php on line 33

tarekwiz 0 Newbie Poster · 2012-06-21T16:06:33+00:00

I Solved It Thanks

rotten69 20 Posting Whiz · 2012-06-21T16:11:20+00:00

Try

    $user = $_POST['usname'];
    $pw = $_POST['psname'];

     if (mysql_num_rows($result) >=1)
     {
         setcookie("user", $user , time()+3600);
         echo ("You Have Successfully Logged In</br>");
     }

If this doesn't work, then unset the cookie and do it again. And,let us know how you go!

tarekwiz 0 Newbie Poster · 2012-06-21T16:20:25+00:00

do you know how to hide some html items with php??

rotten69 20 Posting Whiz · 2012-06-21T17:01:14+00:00

How do you exactly mean? If you're talking about what I am thinking, then you use client-side languages(javascript, jquery)

diafol · 2012-06-21T17:08:36+00:00

Change the password type attribute:

<form action="login.php" method="post">
    <label for = "user">Username: </label><input type="text" name="user" id="user" />
    <label for = "password">Password: </label><input type="password" name="password" id="password" /></br>
<input type="submit" name="sublog" value="Login" />
</form>

Your php:

<?php
$con = mysql_connect("localhost","root","");
mysql_select_db("logindb", $con);
    $user = mysql_real_escape_string($_POST['user']);
    $pw = mysql_real_escape_string($_POST['password']);

    $sql = "SELECT username, password, email FROM users WHERE username='$user' LIMIT 1";
    $result = mysql_query($sql) or die($sql."<br/><br/>".mysql_error());
    if (mysql_num_rows($result)){
         $data = mysql_fetch_assoc($result);
         //IDEALLY you'd hash your pw with a salt - this is not very secure
         if($pw == $data['password']){
             setcookie("user", $user , time()+3600);
             echo "You Have Successfully Logged In<br />";
         }
    }else{
        echo "You Have Failed To Logged In<br />";
    }
?>
<a href="/index.php">CLick here To go back Home</a>

rotten69 20 Posting Whiz · 2012-06-21T17:58:22+00:00

I don't think he is looking that much security at this stage. But, it is a good intro to SQL injection!

tarekwiz 0 Newbie Poster · 2012-06-21T18:20:14+00:00

Thanks all but
there is a problem now i want to add html code to php if-else statements (to show or hide something)
i mean like that but when i use echo like

<?php
echo "<form action="login.php" method="post">
    <label for = "user">Username: </label><input type="text" name="user" id="user" />
    <label for = "password">Password: </label><input type="password" name="password" id="password" /></br>
<input type="submit" name="sublog" value="Login" />
</form>";
include 'menu.php';
?>

it says this error
Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\xampp\htdocs\index.php on line 2

tarekwiz 0 Newbie Poster · 2012-06-21T19:58:35+00:00

up solved but have a new error :S
Here is Code

<?php
$cookie = $_COOKIE["userforcong"];
if (isset($_COOKIE["user"]))
            {
            echo '</br>Congratulations You have registered And logged in with your account ( <?php echo "$cookie"; ?> ).</br>';
                echo '<form action="" method="post">
                        <input type="submit" name="logout" value="Logout" />
                        </form>';


            }
            else
            {
                echo 'Please Login And Come back again';
            }




    function logout()
    {

        if (isset($_COOKIE["user"]))
            {
                setcookie("user", "", time()-3600);
                echo '</br>Logout Successfully</br>';
            }

    }

if (isset($_POST['logout']))

{
     logout();
}

?>

<html>
</html>

But i Get this error
Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\cong.php:5) in C:\xampp\htdocs\cong.php on line 25

diafol · 2012-06-21T20:43:53+00:00

Try:

echo "<br />Congratulations You have registered And logged in with your account ($cookie)<br />";

Also, you should check for the existence of a superglobal ($_COOKIE) before ascribing it to another var:

$cookie = $_COOKIE["userforcong"];

This will end in tears if $_COOKIE["userforcong"] does not exist.

tarekwiz 0 Newbie Poster · 2012-06-21T20:59:50+00:00

i made COOKIE userforcong
im sure it exist

tarekwiz 0 Newbie Poster · 2012-06-21T21:03:00+00:00

and still same error

diafol · 2012-06-21T21:17:52+00:00

Ah! I was looking at line 5 in the error message - you're problem is at line 25:

function logout()
    {
        if (isset($_COOKIE["user"]))
            {
                setcookie("user", "", time()-3600);
                echo '</br>Logout Successfully</br>';
            }
    }

You can't set a cookie once you've echoed stuff to the screen or have got any html/whitespace in the page (http://php.net/manual/en/function.setcookie.php). I'm assuming that your page is included inside another file? This file seems to have output before the include statement. I may be mistaken.

tarekwiz 0 Newbie Poster · 2012-06-21T21:20:32+00:00

when i tried to type in 
`           echo ('hey.'.$cookie.'');`
            it gave me a good response idk why it doesnt wanna work
            P.s It Works With the successfull statement (after edit)

tarekwiz 0 Newbie Poster · 2012-06-21T21:39:27+00:00

its not that problem i have echoed another cookie it didnt touch user cookie

tarekwiz 0 Newbie Poster · 2012-06-21T21:44:58+00:00

i made it i just made the button to redirect to a page that only have this code and it worked

<?php
    function logout()
    {

        if (isset($_COOKIE["user"]))
            {
                setcookie("user", "", time()-3600);
                echo '</br>Logout Successfully</br>';
            }

    }
    logout();
?>

diafol · 2012-06-21T22:28:13+00:00

Great. Solved?