can't sign in to mysql using posted $user and $pass1

[gloosemo]

this code is to create a new user based on posted variables. it works fine for creating the necessary tables and granting privileges. but then when i log out as root and try to log back in as the user it's failing. why is it failing? Thanks, G.

 

		<?php
			extract( $_POST );
			include( '../php/userobject.php' );

			@ $db = mysqli_connect('localhost', 'root', 'xxxxxx', 'mealchamp');
			if ( mysqli_connect_errno() ) {
				echo 'Error: could not connect to database as root.<br />';
				exit;
			}
			else
				echo 'Connected to database as root.<br />';

			$query = "SELECT username FROM userlist WHERE username = $uname";
			$result = mysqli_query($db, $query);

			if ($result) 
				echo 'That user name is already taken.<br/>Please hit back on your browser, and try a different user name.<br/>';
			else {
				echo 'That username is available.<br/>';
				echo "$fname<br/>";
				echo "$lname<br/>";
				echo "$add1<br/>";
				echo "$add2<br/>";
				echo "$city<br/>";
				echo "$prov<br/>";
				echo "$ctry<br/>";
				echo "$post<br/>";
				echo "$email<br/>";
				echo "$uname<br/>";
				echo "$pass1<br/>";
				echo "$sex<br/>";
				echo "$birth<br/>";
				echo "$height<br/>";
				echo "$weight<br/>";

				$query = "INSERT INTO userlist 
						(firstname, lastname, address1, address2, city, 	province, country, postal, email, 	username, password, sex, 		birth, 		height, 		weight) VALUES
						('$fname',	'$lname',	'$add1',	'$add2',	'$city','$prov',	'$ctry', '$post','$email','$uname',	'$pass1',	'$sex', '$birth', '$height', 	'$weight')";
				$result = mysqli_query($db, $query);
				if ($result)
					echo "Userlist was updated successfully<br />";
				else
					echo "Could not update userlist";

				$query = "CREATE TABLE ".$uname."_persfood  
					(persfoodid 	INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
					 food			 		VARCHAR(30) NOT NULL)";
				$result = mysqli_query($db, $query);
				if ($result)
					echo "Personal food table was created successfully<br />";
				else
					echo "Could not create personal food table<br />";

				$query = "GRANT select ON mealchamp.userlist TO ".$uname." IDENTIFIED BY '$pass1'";
				$result = mysqli_query($db, $query);
				if ($result)
					echo "User priveleges were granted successfully<br />";
				else
					echo "Could not grant user priveleges<br />";
				$query = "GRANT select, insert, update, delete, index ON mealchamp.".$uname."_persfood TO ".$uname." IDENTIFIED BY '$pass1'";
				$result = mysqli_query($db, $query);
				if ($result)
					echo "User priveleges were granted successfully<br />";
				else
					echo "Could not grant user priveleges<br />";

				mysqli_close($db);

				@ $db = mysqli_connect('localhost', $uname, $pass1, 'mealchamp');
				if ( mysqli_connect_errno() ) {
					echo 'Error: could not connect to database as user.<br />';
					exit;
				}
				else {
					echo 'Connected to database as user.<br />';
?>

[Pikachu2000]

In what way is it failing?

[gloosemo]

it says at the end the mysqli_connect_errno() message ---> echo 'Error: could not connect to database as user.'

 

Then when i try to do things with the $db resource i get warnings that it is not a valid resource. The $db connection is not being established. Did i miss quotation marks somewhere or something???

 

thanks again, G

[Pikachu2000]

Instead of echoing a generic error message for development, show the actual error message returned by MySQL.

 

if ( mysqli_connect_errno() ) {
echo '<br>Connection error: ' . mysqli_connect_error() . '<br />';
etc . . . 

[gloosemo]

Error: Access denied for user 'pleasework'@'localhost' (using password: YES)

 

that is when

 

$uname = pleasework

$pass = pleasework

 

and this time i queried "FLUSH PRIVILEGES" after doing the grants.

 

does it have something to do with maybe i should be adding @localhost to the usernames somewhere?