whats wrong? - i know i missed some simple thing but well help please :)

codefreek · 06-26-2008, 05:31 PM

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PHP Code:


			
<?php
error_reporting(E_ALL & ~E_NOTICE);
include("db_connect.php");

$sql = "SELECT * FROM `users` WHERE `id` = '".$id."'";
                $query = mysql_query($sql) or die(mysql_error());
        while($sel_table = mysql_fetch_array($query)) 
        {
  print $sel_table['username'];

}
    $net = "UPDATE `users` SET `username` = '".$username."', `password` = '".$password."' WHERE `id` = '".$id."'";
    $web_result = mysql_query($net) or die(mysql_error());


?>

no errors are made i am trying to do print $sel_table['username'] or even post the id or something but for some reason it do not print..

so please tell me whats wrong :S

//Code,-No flame please ;)-

Code:

CREATE TABLE `users` (
  `id` int(11) NOT NULL auto_increment,
  `username` varchar(30) NOT NULL,
  `password` varchar(40) NOT NULL,
  `email` text NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

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delayedinsanity · 06-26-2008, 06:17 PM

Where are you getting $id from? And there's nothing going on inside of your while loop so I'm not sure where you're getting the other variables from either, but if $id is an empty value then chances are your query isn't returning any errors, but it's not returning any data either.
-m

codefreek · 06-26-2008, 07:06 PM

Quote:

Originally Posted by delayedinsanity

Where are you getting $id from? And there's nothing going on inside of your while loop so I'm not sure where you're getting the other variables from either, but if $id is an empty value then chances are your query isn't returning any errors, but it's not returning any data either.
-m

//delayedinsanity
I just want to know why it is not letting me get any of the data from db table users?..

like $sel_table['username']
or $username as a var,

Thank you in adv.

EDIT:
i get bool(false) bool(true)
on

Code:

var_dump($sel_table);
var_dump($web_result);

EDIT:

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delayedinsanity · 06-26-2008, 07:20 PM

That's what I'm trying to do as well.

Which is why I was wondering if maybe that was an incomplete code segment and perhaps the problem was located elsewhere, since I can't see anywhere that $id is being defined, and if you're searching for an empty ID, or an ID that doesn't exist in the table, then you're going to get an empty result which would be why $sel_table isn't receiving any data.
-m

codefreek · 06-26-2008, 08:49 PM

That is the code segment, i didn't code anything else yet because of the fact that the code is not working

EDIT:

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codefreek · 06-26-2008, 09:30 PM

i fixed the problem thank you ;)

New problem look down...

EDIT:

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Jim · 06-26-2008, 09:32 PM

Ok to make it damn easy, echo the $sql variable. I bet we can learn much more from that :)

I go with delayedinsanity, you cannot use variables that don't exist (yet). You first have to create the variables that you want to be used. Both $id and $username don't exist, so they will return NULL. (NULL = basically _nothing_)

Edit// Damn too late :(

codefreek · 06-26-2008, 09:42 PM

PHP Code:


			
<?php
error_reporting(E_ALL & ~E_NOTICE);
include("db_connect.php");
session_start(); // Starts the session.

if ($_SESSION[&#8216;logged’] == 1) { // User is already logged in.

        header("Location: index.php"); // Goes to main page.

        exit(); // Stops the rest of the script.

} else {

        if (!isset($_POST[&#8217;submit’])) { // The form has not been submitted.

echo" 
<form action='users.php' name='login' method='post'> 
<table> 
    <tr> 
        <td> 
        username 
        </td> 
    </tr> 
    <tr> 
        <td> 
        <input type='text' name='$username'> 
        </td> 
    </tr> 

        <tr> 
        <td> 
        password
        </td> 
    </tr> 
    <tr> 
        <td> 
        <input type='text' name='$password'> 
        </td> 
    </tr> 
    <tr> 
        <td> 
        
        </td> 
    </tr> 
    <tr> 
    </tr> 
    <tr> 
        <td> 
        <input type='submit' name='login' value='login'> 
        </td> 
    </tr> 
</table> 
</form>";     
    
    } else {

                $username = form($_POST[&#8216;username’]);

                $password = md5($_POST[&#8216;password’]); // Encrypts the password.

               

                $q = mysql_query("SELECT * FROM `users` WHERE username = ‘$username’ AND password = ‘$password’") or die (mysql_error()); // mySQL query

                $r = mysql_num_rows($q); // Checks to see if anything is in the db.

               

                if ($r == 1) { // There is something in the db. The username/password match up.

                        $_SESSION[&#8216;logged’] = 1; // Sets the session.

                        header("Location: index.php"); // Goes to main page.

                        exit(); // Stops the rest of the script.

                } else { // Invalid username/password.

                        exit("Incorrect username/password!"); // Stops the script with an error message.

                }

        }

}

?>

no errors but the login do not work :S

codefreek · 06-26-2008, 09:49 PM

if you see #8216 in the form :S it is the forum that is making that not me :S
here is code on
PasteBin.be
pastebin if some one wants to take a look at the script in a more clean space,

delayedinsanity · 06-26-2008, 10:03 PM

Your form might be causing the problem. You need the fields to look something more like

PHP Code:


			
echo '<input type="text" name="username" value="' . $username . '" />';
echo '<input type="text" name="password" value="' . $password . '" />';

There is nothing in $_POST['username'] or $_POST['password'] because you're using variables to determine the name of those fields. So for all we know, the information could be showing up in $_POST['codefreek']. More likely it's going nowhere, as the name field is probably empty when the page is displayed.

Might I suggest using CSS to style your forms too?

-m

codefreek · 06-26-2008, 10:30 PM

thank you i don't style my applications before everything is done..

codefreek · 06-26-2008, 10:33 PM

so should i take off the $_POST?

codefreek · 06-26-2008, 10:43 PM

Quote:

Originally Posted by delayedinsanity;

Your form might be causing the problem. You need the fields to look something more like

PHP Code:


			
echo '<input type="text" name="username" value="' . $username . '" />';
echo '<input type="text" name="password" value="' . $password . '" />';

how do i put this in my form ?
please reform it to, put it in my form # to match my form..
Ty.

NVM I FIXED ;)

codefreek · 06-26-2008, 10:49 PM

FORM EDIT:

PHP Code:


			
      
                echo "<form action=\"users.php\" method=\"POST\">";

                echo "<table>";

                echo "<tr>";

                echo "<td colspan=\"2\">Login:</td>";

                echo "</tr>";

                echo "<tr>";

                echo '<input type=\"text\" name=\"username\" value="'.$username.'">';

                echo "</tr>";

                echo "<tr>";

                echo '<input type=\"text\" name=\"password\" value="'.$password.'">';

                echo "</tr>";

                echo "<tr>";

                echo "<td colspan=\"2\"><input type=\"submit\" name=\"submit\" value=\"submit\"</td>";

                echo "</tr>";

                echo "</table>";

                echo "</form>";

codefreek · 06-26-2008, 10:53 PM

Some Thing is still wrong, i know everything should work now but it do not work!

delayedinsanity · 06-26-2008, 11:20 PM

When you're using single quotes (') you don't have to escape double quotes (") and visa versa. This isn't how I would do it normally, but to fit with the style of what you're doing;

PHP Code:


			
if ( ! isset($username))
{
    $username = '';
}

if ( ! isset($password))
{
    $password = '';
}

$szForm = <<<FORM
<form action="users.php" name="login" method="post"> 
<table> 
    <tr><td>username</td>
        <td><input type="text" name="username" value="{$username}" /></td>
    </tr> 

    <tr><td>password</td>
        <td><input type="password" name="password" value="{$password}" /></td>
    </tr>

    <tr><td colspan="2"><input type='submit' name='login' value='login' /></td></tr> 
</table> 
</form>
FORM;

echo $szForm;

Typos and simple mistakes can happen to anybody at anytime, so never be afraid to view the source directly from your browser, or echo any variables to the screen temporarily to check them, and so on and so forth.
-m

codefreek · 06-26-2008, 11:34 PM

Thank you delayedinsanity for all the help :D
wont forget you ;D

EDIT:
Fatal error: Call to undefined function form() in line 45

FIXED THAT PROBLEM

codefreek · 06-26-2008, 11:42 PM

is there any reference on all the errors because i have no idea what that error stands for :S

codefreek · 06-26-2008, 11:53 PM

now, i got a new problem,
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '‘ AND password = ‘' at line 1

PHP Code:


			

$q = mysql_query("SELECT * FROM `users` WHERE username = ‘$username’ AND password = ‘$password’") or die (mysql_error()); // mySQL query

i tried to change it to

PHP Code:


			
'".$rest."'

but that didn't work

I FIXED THAT PROBLEM

codefreek · 06-27-2008, 12:18 AM

now i have a new problem i login with right pass and username, it all works sends me to index if the user is logged in but when i try to logout..
it wont logout :S

THIS IS ON index:

PHP Code:


			
session_start(); // Starts the session.



if ($_SESSION[‘logged’] != 1) { // There was no session found!



        header("Location: users.php"); // Goes to login page.



        exit(); // Stops the rest of the script.



}



echo "This is the main page!";

LOGOUT page,

PHP Code:


			
<?php



      session_unset(); // Destroys the session.



      header("Location: users.php"); // Goes back to login.





?>

and this is the users page

PHP Code:


			
<?php

error_reporting(E_ALL & ~E_NOTICE);

include("db_connect.php");

session_start(); // Starts the session.



if ($_SESSION[‘logged’] == 1) { // User is already logged in.



        header("Location: index.php"); // Goes to main page.



        exit(); // Stops the rest of the script.



} else {



if ( ! isset($username))

{

    $username = '';

}



if ( ! isset($password))

{

    $password = '';

}



$szForm = <<<FORM

<form action="users.php" name="login" method="post"> 

<table> 

    <tr><td>username</td>

        <td><input type="text" name="username" value="{$username}" /></td>

    </tr> 



    <tr><td>password</td>

        <td><input type="password" name="password" value="{$password}" /></td>

    </tr>



    <tr><td colspan="2"><input type='submit' name='login' value='login' /></td></tr> 

</table> 

</form>

FORM;



echo $szForm;  

   

    





               $password = mysql_real_escape_string($_POST['password']);

           $username = mysql_real_escape_string($_POST['username']);









               



                $q = mysql_query("SELECT * FROM users WHERE username = '$username'

                 AND password = '$password'") or die (mysql_error()); // mySQL query



                $r = mysql_num_rows($q); // Checks to see if anything is in the db.



               



                if ($r == 1) { // There is something in the db. The username/password match up.



                        $_SESSION[‘logged’] = 1; // Sets the session.



                        header("Location: index.php"); // Goes to main page.



                        exit(); // Stops the rest of the script.



                } else { // Invalid username/password.



                        exit("Incorrect username/password!"); // Stops the script with an error message.



                }



        }







?>