Jump to content
Stef's Coding Community

jamie

Member
  • Content Count

    9
  • Joined

  • Last visited

Community Reputation

-1 Poor

About jamie

  • Rank
    New member
  1. I cannot get logged in to the system. I deleted the admin because it wasn't working and I inserted a new user using phpmyadmin and I still get the error when trying to login. How can I make a new user so that I can view the members.php page
  2. I forgot about the temp.php. I copied it exactly so I also have jk7d?3 but when I put that into the password box, I still get the error. I am sure it is not your code. maybe I will just make a new admin and that should fix it.
  3. I am having trouble logging in to the members.php with my code. I thought the password was set up in the config with this line $config['salt'] = 'jk7d?3'; so when I try to login, I keep getting the error. Of course the password field in the database just shows a bunch of characters
  4. okay, that sounds good. I am going to follow through the tutorial and make the login system. It would be awesome if you could explain things a bit more in future video tutorials. I am trying to learn php and I am going to try out this site to see how it works compared to other sites. I have some recommendations if you are interested to improve your site here P.S, ben I really like your website. It looks great. Did you ever get that company to pay you for stealing your design?
  5. I am following one of the video tutorials but the author isn't explaining the code fully. So I am looking to understand why he is coding it in this way. <form action="" method="post"> <div> <?php if($error['alert'] != ''){ echo "<div class='alert'>" . $error['alert'] . "</div>"; } ?> <label for="username">Username: *</label> <input type="text" name="username" value="<?php echo $input['user']; ?>" /> <div class="error"><?php echo $error['user']; ?></div> <label for="password">Password: *</lab
  6. Never mind, I got it figured out. I left out an I on the MYSQL
  7. I am following one of the video tutorials - making a php mysql login system by ben faulk and I am getting this error on the login.php Notice: Use of undefined constant MYSQL_REPORT_ERROR - assumed 'MYSQL_REPORT_ERROR' in C:\wamp\www\phpLogin\includes\config.php on line 13 Warning: mysqli_report() expects parameter 1 to be long, string given in C:\wamp\www\phpLogin\includes\config.php on line 13 here is the config.php //user authentication $config['salt'] = 'jk7d?3'; //error reporting mysqli_report(MYSQL_REPORT_ERROR); ?>
×
×
  • Create New...