pop up message did not showed up

Question

eyeda 0 Newbie Poster

13 Years Ago

Anybody can help me with this code? Why the pop up message for deactivated account did not showed up?

<?
include("database.php");

$myid=$_POST['myid']; 
$mypassword=$_POST['mypassword']; 
$status=$_POST['mystatus'];


if ($_POST['user']='Admin'){
     $resultad = mysql_query("SELECT * FROM staff WHERE staff_id='$myid' AND level='ADMIN' AND password='$mypassword'" );
     $count1 = mysql_num_rows($resultad);
     $row1 = mysql_fetch_array($resultad);


                    if($count1==1){ 


                    header('Location:adLandingPage.php');

                    $insertSQL = "INSERT INTO login_table (id,time) VALUES ('$myid',NOW())";
                    $Result1 = mysql_query($insertSQL) or die(mysql_error());

                            }
                    if ($count1==1 && $row1['enabled'] =="NOT ACTIVE") 
    {
       session_unset();
echo "<script type='text/javascript'>alert('Your account have been deactivated.Please contact your admin!');window.location.href='Index.html'; </script>";
    }

                        }   


if ($_POST['user']='Staff'){

            $result = mysql_query("SELECT * FROM staff WHERE  level='STAFF' AND staff_id='$myid' AND password='$mypassword' AND enabled='ACTIVE' ");
            $result2 = mysql_query("SELECT * FROM staff WHERE staff_id='$myid' AND password='$mypassword' ");


            $count=mysql_num_rows($result);
            $row = mysql_fetch_array($result2);
            $count2=mysql_num_rows($result2);

                    if($count==1){ 
                    session_start();
                    session_register('id');
                    session_register ('password');
                    header('Location:LandingPage.php');

                    $insertSQL = "INSERT INTO login_table (id,time) VALUES ('$myid',NOW())";
                    mysql_select_db($db_name);
                    $Result1 = mysql_query($insertSQL) or die(mysql_error());

                    }



                }




if($myid==""){

session_unset();
echo ' <script type="text/javascript">alert("Please enter your ID!");history.go(-1);
    </script>';}

if($mypassword==""){

session_unset();
echo ' <script type="text/javascript">alert("Please enter your password");history.go(-1);
    </script>';}

if ($count2==0)
    {
        session_unset();
echo ' <script type="text/javascript">alert("Wrong Login!");window.location.href="Index.html"; 
    </script>';
    }
if ($count2==1 && $row["enabled"] == "NOT ACTIVE") // This equates to 'if there is 1 record AND the active field equals 0 (false)'
    {
       session_unset();
echo "<script type='text/javascript'>alert('Your account have been deactivated.Please contact your admin!');window.location.href='Index.html'; </script>";
    }





?>

javascript php

3 Contributors
3 Replies
129 Views
1 Hour Discussion Span
Latest Post 13 Years Ago Latest Post by eyeda

Dani AI

Generated 9 Months Ago

Two things in the original thread explain why the deactivated-account alert never appeared: an assignment-instead-of-comparison bug (reported by ) made the user-branch checks behave wrongly, and the script sent a Location redirect before the alert code ran so the browser never showed the JavaScript. ’s question about using different row variables was also useful — multiple queries/posts can be correct, but they must be read and used in the right order.

Practical fixes and checklist:

Start the session at the very top and use $_SESSION (avoid deprecated session_register). Call session_start() before session_unset() or inspecting session vars.
Validate and trim POST inputs before querying. Handle empty ID/password early to avoid unnecessary DB work.
Query once for the account, then decide: if no row => wrong login; if enabled != 'ACTIVE' => set a server-side message and redirect; else verify password and set session values and redirect. Check for enabled before issuing any success redirect.
Never mix header('Location: ...') and later expect client-side alerts to run; use one method (server-side flash messages are more reliable). Always follow header() with exit to stop further script execution.
Replace old mysql_* code with PDO or mysqli and use prepared statements; store passwords hashed with password_hash/password_verify.

Example flow (PDO, prepared statements, flash-style redirects):

<?php
session_start();
$pdo = new PDO($dsn, $user, $pass);
$stmt = $pdo->prepare('SELECT staff_id,level,enabled,password FROM staff WHERE staff_id = ?');
$stmt->execute([$myid]);
$row = $stmt->fetch(PDO::FETCH_ASSOC);

if (!$row) { $_SESSION['flash']='Wrong Login'; header('Location:Index.html'); exit; }
if ($row['enabled'] !== 'ACTIVE') { $_SESSION['flash']='Account deactivated'; header('Location:Index.html'); exit; }
if (!password_verify($mypassword, $row['password'])) { $_SESSION['flash']='Wrong Login'; header('Location:Index.html'); exit; }

$_SESSION['id'] = $row['staff_id'];
header($row['level']==='ADMIN' ? 'Location:adLandingPage.php' : 'Location:LandingPage.php');
exit;
?>

Extra notes: enable detailed error reporting while debugging (temporarily), log DB errors, and avoid revealing internal messages to users. The arrangement and order of checks matter most — reordering to validate enabled status before any successful redirect fixed the original problem for .

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Member #949455

13 Years Ago

Anybody can help me with this code? Why the pop up message for deactivated account did not showed up?

You have 2 lines that has that code.

Line 24 to 28

 if ($count1==1 && $row1['enabled'] =="NOT ACTIVE")
{
session_unset();
echo "<script type='text/javascript'>alert('Your account have been deactivated.Please contact your admin!');window.location.href='Index.html'; </script>";
}

Line 80 to 84

if ($count2==1 && $row["enabled"] == "NOT ACTIVE") // This equates to 'if there is 1 record AND the active field equals 0 (false)'
{
session_unset();
echo "<script type='text/javascript'>alert('Your account have been deactivated.Please contact your admin!');window.location.href='Index.html'; </script>";

May I ask what is $row1[ ]? Shouldn't be just $row[ ] for both examples?

Edited 13 Years Ago by Member #949455 because: grammer

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GliderPilot 33 Posting Whiz · Answer 1 · 2012-12-06T03:43:27+00:00

He has three different DB queries, one is getting admins ($row1) one is geting staff with a banned account ($row) and the other gets Staff withour a banned account (no variable set for this one). While this method works it would be just as easy to make this all one DB query to reduce the number of DB queries you need to make (And don't forget we shouldn't be using mysql_ functions anymore)

All that asside your IF statements on lines 9 and 33 are assigning variables to your POST data rather than checking them as you only have single statements

9  if ($_POST['user']='Admin'){
33 if ($_POST['user']='Staff'){

should be

9  if ($_POST['user']=='Admin'){
33 if ($_POST['user']=='Staff'){

eyeda 0 Newbie Poster · Answer 2 · 2012-12-06T04:00:52+00:00

@LastMitch
i used $row[] and $row1[] to differentiate the mysQL coding.

thanks a lot, i didn't notice that

Thanks guys,
i've solve it =) it just because of the arrangement of the coding make it hard to read and that make the system didnt work

pop up message did not showed up

Dani AI

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