[php login video tutorial problem]

sent

[php login video tutorial problem]

I will do that right now

[php login video tutorial problem]

You put "admin" into the password box?

 

According to the way I set things up by default, the username is "admin" and the password is "admin" (obviously you would want to change this to something more secure on a live site.)

 

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

[php login video tutorial problem]

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.

[php login video tutorial problem]

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

[why code it this way?]

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?

[why code it this way?]

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: *</label>
		<input type="password" name="password" value="<?php echo $input['pass']; ?>" />
		<div class="error"><?php echo $error['pass']; ?></div>
		<p class="required"> * required fields</p>

		<input type="submit" name="submit" class="submit" value="submit" />
	</div>
</form>

 

The question here is why use the $error['alert']? this is an array. I don't see where the $error[] comes from. The only place it is defined is in another file and it is defined as nothing

$error['alert'] = '';

 

so what I am trying to figure out is why do it this way and not this way

 

$error = '';

[getting an error from one of the videos]

Never mind, I got it figured out. I left out an I on the MYSQL

[getting an error from one of the videos]

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);
?>