Creating a Login System in PHP using Oracle

Question

I'm having massive issues with creating this login system for my website and we are required to use php and oracle.

The table itself is very simple and only has a Username and Password value attached to it.

This is the code I am using and the main issue that comes with it is that the variable $password always returns a blank value.

<?php
/* Set oracle user login and password info */
$dbuser = "*MY USERNAME*";
$dbpass = "*MY PASSWORD*";
$dbname = "SSID";
$db = oci_connect($dbuser, $dbpass, $dbname);

if (!$db)  {
    echo "An error occurred connecting to the database"; 
    exit; 
}

$user = $_POST['user'];
$pass = $_POST['pass'];

$sql_login = "SELECT Username FROM users WHERE Username='%".$user."%'"; 

$login_stmt = oci_parse($db, $sql_login);

if(!$login_stmt)
{
    echo "An error occurred in parsing the sql string.\n"; 
    exit; 
}

oci_execute($login_stmt);

while(oci_fetch_array($login_stmt))
{
    $password = oci_result($login_stmt,"Password");
}

if ($password == "")
{
    echo 'Password = blank';
}

if ($pass == $password)
{
    echo 'Logged In';
}
else
{
    echo 'Login Failed';
}


?>

I am using this command to try and write a value to the variable but I am having no luck.

while(oci_fetch_array($login_stmt))
{
    $password = oci_result($login_stmt,"Password");
}

The form used is below, but I don't think there is a problem with it.

<form name="register" method="post" action="inc/login.php">          
                <div class="form_row">
                <label class="contact"><strong>Username:</strong></label>
                <input type="text" name="user" class="contact_input" />
                </div>  


                <div class="form_row">
                <label class="contact"><strong>Password:</strong></label>
                <input type="password" name="pass" class="contact_input" />
                </div>                     

                <div class="form_row">
                    <div class="terms">
                    <input type="checkbox" name="terms" />
                    Remember me
                    </div>
                </div> 


                <div class="form_row">
                <input type="submit" class="register" value="login" />
                </div>   
</form>

problem could be in your form itself; you should post it. Use error reporting in the meantime to catch and display http://php.net/manual/en/function.error-reporting.php — Funk Forty Niner, Oct 01 '15 at 12:37
I tried both, but the problem I am having is assigning a value to $password. Unless there is an easier way to check whether the input password is the same as the one stored in the database — user2854076, Oct 01 '15 at 12:43
thats what this is for, except it is returning the varible as blank while(oci_fetch_array($login_stmt)) { $password = oci_result($login_stmt,"Password"); } — user2854076, Oct 01 '15 at 12:45
Change your query to `SELECT Username, Password FROM users WHERE Username='%".$user."%'` — Epodax, Oct 01 '15 at 12:46
you're using LIKE syntax `WHERE Username='%".$user."%'` which should technically read as `WHERE Username LIKE '%".$user."%'`. But you want to check for truthness and not likeness `WHERE Username='".$user."'` — Funk Forty Niner, Oct 01 '15 at 12:59
The problem wasn't finding the password. I had to select all of the values from the database instead of the one and sort through them afterwards. The problem is solved now. — user2854076, Oct 01 '15 at 13:03
great, either post your own answer or you can delete the question. Far as I'm concerned, the answer given below should have been posted as a comment. Edit: they decided to delete their "answer". — Funk Forty Niner, Oct 01 '15 at 13:04

score 0 · Answer 1 · edited May 23 '17 at 12:07

0

problem is sql return 0 rows, bacase if u using in where clause % U must use like operator, not =

use this sql:

$sql_login = "SELECT Username FROM users WHERE Username like '%".$user."%'";

Here is good informatioun about this:

Equals(=) vs. LIKE

edited May 23 '17 at 12:07

Community

1
1

answered Apr 28 '16 at 20:48

Michal Hože

116
4

score 0 · Answer 2 · answered Jul 02 '17 at 21:02

0

In $password = oci_result($login_stmt,"Password"); the word "Password" must be written on CAPS so it will be something like

$password = oci_result($login_stmt,"PASSWORD");

Worked that way for me

answered Jul 02 '17 at 21:02

Ricardo Nuñez

1

Creating a Login System in PHP using Oracle

2 Answers2