使用PHP和MYSQL登录页​​面基于用户角色

Question

I created a login page.

I want customer and manager to login from this same page. I separated customer and manager in db by their role "","manager" respectively (customer represented as null value).

I wrote this code but its not working as i expected. When I type name and password and press login button it doesn't go to desired page rather it remains on login page.

<?php
session_start();

$db=mysqli_connect("localhost","root","","authentication");

if (isset($_POST['login_btn'])) {

    $customer_name = mysql_real_escape_string($_POST['customer_name']);
    $password = mysql_real_escape_string($_POST['password']);
    $password=md5($password);
    if(empty($customer_name)&&empty($password)) {
        $error="Fields are Mandatory";
    } else {
        $sql="SELECT * 
              FROM customers 
              WHERE customer_name='$customer_name' AND password='$password'";
        $result=mysqli_query($db,$sql);
        $row=mysqli_fetch_array($result);
        $_SESSION['id']=$row['id'];
        $_SESSION['role']=$row['role'];
        $count=mysqli_num_rows($result);

        if($count==1){
            if ($row['role']=="manager"){
                header("location:manager.php");      
            }
            else if($row['role']==""){
                header("location:ueser.php");      
            }
        }else{
            $error='Customer_name/password combination incorrect';
        }
    }
}
?>

Answer 1

Use elseif instead else if

/* Incorrect Method: */
if ($a > $b):
    echo $a." is greater than ".$b;
else if ($a == $b): // Will not compile.
    echo "The above line causes a parse error.";
endif;


/* Correct Method: */
if ($a > $b):
    echo $a." is greater than ".$b;
elseif ($a == $b): // Note the combination of the words.
    echo $a." equals ".$b;
else:
    echo $a." is neither greater than or equal to ".$b;
endif;

For More http://php.net/manual/en/control-structures.elseif.php#control-structures.elseif