如果while循环中的if语句不起作用

Question

This is a login php code that allows different users to login and use the application

$con=mysql_connect("localhost","root","");

mysql_select_db("sg",$con);


if(isset($_POST['uid'])){ $username = $_POST['uid']; } 

if(isset($_POST['pwd'])){ $password = $_POST['pwd']; } 





error_reporting(0);
$result=mysql_query("SELECT username,password from user where username = '$username' and password='$password'");

I want to check if a user exists and then open different pages for different users.In all three cases the page is redirected to Analyst.php

    while($info = mysql_fetch_array( $result ))     

        {

if($username='operations')
{
                 header("Location: view3.php");
}

if($username='finance')
{
         header("Location: view3.php");
}

if($username='Analyst')
{
         header("Location: Analyst.php");
}

}           
 echo "<script type='text/javascript'>alert('Invalid userID or password ');window.location=\"home.html\";</script>";    


 ?>

Answer 1

You are assigning values to variable $username instead of comparing the variable to string. Try with :

if($username=='operations')
{
   header("Location: view3.php");
}

if($username=='finance')
{
   header("Location: view3.php");
}

if($username=='Analyst')
{
   header("Location: Analyst.php");
}

Answer 2

You need to use two = signs:

if($username == 'finance')

A single = will assign a value to $username . In order to assess the value, you should use == .

Better still, you could use a switch() statement:

while($info = mysql_fetch_array( $result ))     
{
    switch($username)
    { 
        case 'operations':
        case 'finnce':
            header("Location: view3.php");
            break;

        case 'Analyst':
            header("Location: Analyst.php");
            break;
    }
}           

echo "<script type='text/javascript'>alert('Invalid userID or password ');window.location=\"home.html\";</script>";    

Please also be advised that the mysql_* family of functions is now deprecated, and should not be used in new code. You should instead look at MySQLi or PDO .