Error with PhP Script login_tools - php

Question

I'm coding a basic PhP login script right now to develop later on, I've got as far as the Login_tools.php file but when I login in with the wrong details (to check login_tools is working) however I get this error:
"PHP Syntax Check: Parse error: syntax error, unexpected '$un' (T_VARIABLE) in your code on line 27
$q = "SELECT user_id, username FROM users WHERE username = '$un' AND pass = SHA1('$p')" ;"
<?php
function load($page = 'login.php')
{
$url = 'http://'.$_SERVER['HTTP_HOST'].dirname( $_SERVER['PHP_SELF']);
$url = rtrim( $url, '\');
$url .= '/'.$page ;
header("Location:$url") ;
exit() ;
}
function validate( $dbc, $username =", $pwd=")
{
$errors = array();
if( empty($username))
{ $errors[] = "Enter your username."; }
else
{$un = mysqli_real_escape_string( $dbc, trim( $username));}
if( empty( $pwd))
{ $errors[]= "Enter your password.";}
else
{$p = mysqli_real_escape_string($dbc, trim($pwd)); }
if(empty( $errors))
{
$q = "SELECT user_id, username FROM users WHERE username = '$un' AND pass = SHA1('$p')" ;
$r = mysqli_query ( $dbc, $q) ;
if(mysqli_num_rows( $r ) == 1)
{
$row = mysqli_fetch_array ($r, MYSQLI_ASSOC) ;
return array (true, $row) ;
}
else
{ $errors[] = 'Username and/or Passsword not found.';}
}
return array ( false, $errors) ; }
?>
Any help as to why I am getting this error would be much appreciated...

Answer 1

The error is in this line:
url = rtrim( $url, '\');
backslash is an escape char.
Change it to
url = rtrim( $url, '\\');
and the error goes away.

Answer 2

Try this one:
<?php
function load($page = 'login.php')
{
$url = 'http://'.$_SERVER['HTTP_HOST'].dirname( $_SERVER['PHP_SELF']);
$url = rtrim( $url, '\\');
$url .= '/'.$page ;
header("Location:$url") ;
exit() ;
}
function validate( $dbc, $username ="", $pwd="")
{
$errors = array();
if( empty($username))
{ $errors[] = "Enter your username."; }
else
{$un = mysqli_real_escape_string( $dbc, trim( $username));}
if( empty( $pwd))
{ $errors[]= "Enter your password.";}
else
{$p = mysqli_real_escape_string($dbc, trim($pwd)); }
if(empty( $errors))
{
$q = "SELECT user_id, username FROM users WHERE username = '$un' AND pass = SHA1('$p')" ;
$r = mysqli_query ( $dbc, $q) ;
if(mysqli_num_rows( $r ) == 1)
{
$row = mysqli_fetch_array ($r, MYSQLI_ASSOC) ;
return array (true, $row) ;
}
else
{ $errors[] = 'Username and/or Passsword not found.';}
}
return array ( false, $errors) ; }
I hope this helps.