Jump to content

Archived

This topic is now archived and is closed to further replies.

dual_alliance

User not found.....

Recommended Posts

I have the following code:

[code]<?php
session_start();
header("Cache-control: private");

// Get MySQL Database information

include('db.php');

// Make variables from the form

$Username = $_POST['userName'];
$Password = $_POST['passWord'];

// Remove HTML (if any)

$Username = strip_tags("$Username");
$Password = strip_tags("$Password");

// Remove escaped characters (if any)

$Username = stripslashes("$Usermame");
$Password = stripslashes("$Password");

// Connect to server and select database.

mysql_connect("$dbHost", "$dbUserName", "$dbPassWord")or die("Cannot connect to server!");
mysql_select_db("$dbName")or die("Cannot select Database!");

// Does the user exist?

$sql_user_check = 'SELECT * FROM users WHERE username="$Username"';
$result_name_check = mysql_query($sql_user_check);
$usersfound = mysql_num_rows($result_name_check);

// If the user doesn't exist, create error

if ($usersfound < 1) {
$error = "User $Username not found.";

// If the user does exist, continue with processing

}else{

// Check if the passwords match

    $sql_pass_get = 'SELECT * FROM users WHERE username="$Username"';
    $user_info = mysql_fetch_array(mysql_query($sql_pass_get));
  $encryptpass = $user_info['password'];

// If it doesn't match, note that and end
  if ($encryptpass != md5($PassWord)) {
    $error = "Invalid password.  Try again.";

// If it does match, let in and pass on info to session variables

}else{

        $_SESSION['userid'] = $user_info['userid'];
      $_SESSION['username'] = $user_info['username'];
      $_SESSION['password'] = $user_info['password'];

    }
}

if (!$_SESSION['username']) {
 
echo "$error";
   
}else{
include('admin.php');

}

?>[/code]

On index.php (login) l enter the username l entered into the MySQL database and it says user not found... I have gone into the MySQL database using phpmyadmin and the username is there.  So what is going on?

Thanks,

dual_alliance

Share this post


Link to post
Share on other sites
$sql_user_check = 'SELECT * FROM users WHERE username="$Username"';

$username is taken as literal, as its in single quotes on the PHP-level

try

$sql_user_check = "SELECT * FROM users WHERE username=\"$Username\"";

Share this post


Link to post
Share on other sites

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.