multiple level login form not working.. while submitting

Question

dont know why this form is not working.... while i submit this after putting username and password.. it donot show any thing after submission

<?php 
 include "connect_to_mysql.php";
 if(isset($_POST['log']))
 {
 $user= $_POST['user'];
 $pass=md5($_POST['pass']);

   $sql=mysql_query("select* from login where user= '$user' AND pass='$pass'    LIMIT 3 ");
  $data=mysql_fetch_array($sql);

  $UserName= $data['user'];   
  $Password= $data['pass'];
  $type= $data['type'];
  $name= $data['name'];

  if($user==$username && $pass==$password){
  session_start();
  $_SESSION['name']=$name;
  if($type=='admin')
  {
         header("location: index.php");
  }
  else if($type=='vender1')
 {
         header("location: vender1.php");
  }

 }




  else {
        echo 'That information is incorrect, try again <a href="index.php">Click Here</a>';
    exit();
  }
  }
  ?>

database is created manually and login table has 5 columns named as id, name. user, pass, type.

     <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
   <head>
   <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
   <title> Log In </title>
   <link rel="stylesheet" href="style.css" type="text/css" media="screen" />
   </head>

   <body>
   <br /><br /><br /><br /><br /><br /><br /><br />
   <br /><br /><br /><br />
   <div id="mainWrapper">
   <?php include_once("header.php") ?></div>
   <div id="pageContent"><br /><br /><br />
   <div align="right" style="margin-right:24px; color:#FF0000">
   <h2>Please Log In To Manage the Inventary</h2>
   <br /><br />
   <form id="form" name="form" method="post" action="login.php">
     <h2 style="padding-right:200px;">User Name:</h2>
       <input name="user" type="text" id="user" size="40"   style="height:20px;" />
     <br /><br />
     <h2 style="padding-right:210px;">Password:</h2>
     <input name="pass" type="password" id="pass" size="40" style="height:20px;" />
    <br />
    <br />
    <br />

      <input type="submit" name="button" id="button" value="Log In" />

    </form>
   <p>&nbsp; </p>
   </div>
   <br />
   <br />
   <br />
   </div>

    </div>
     </body>
     </html>

i m stucked here.. please make me out of this

Your code is SQL Injective and please use PDO, you are using deprecated functions like `mysql_query()` http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php — Umair Ayub, Jul 31 '15 at 07:28
I am not sure but maybe you have error in SQL syntax Change `select*` to `select *` — Umair Ayub, Jul 31 '15 at 07:29
i didnt get you.. please write down in code.. what you are saying — deep singh, Jul 31 '15 at 07:30
You can run your query like `mysql_query() or die(mysql_error())` so you can see if your query ran or not — Umair Ayub, Jul 31 '15 at 07:30
`$sql=mysql_query("select * from login where user= '$user' AND pass='$pass' LIMIT 3 ") or die(mysql_error());` change that line to this — Umair Ayub, Jul 31 '15 at 07:31
You have used $UserName for declaration and used $username. Please check your code. :) — Blue Rose, Jul 31 '15 at 07:33
@Umair this is not answer my man.. i have checked.. even this is not working — deep singh, Jul 31 '15 at 07:36
Did you replaced the line to which I have told in my answer, it will show the if you have error in your MYSQL — Umair Ayub, Jul 31 '15 at 07:40

Umair Ayub · Answer 1 · 2015-07-31T07:44:10.137

1

You have error in your query, you are writing select* but there should be a space like so select *

$sql = mysql_query("select * from login where user= '$user' AND pass='$pass' LIMIT 3 ") or die(mysql_error());

EDIT

Also YOU MUST HAVE TO PUT session_start(); as the first line of your code... else it would not work.

So make this as your first lines of code

session_start();
error_reporting(E_ALL); // to see if there is error in code

And also PHP varialble names are case-sensitive,

Variables in PHP are represented by a dollar sign followed by the name of the variable. The variable name is case-sensitive.

So please change

if($user==$username && $pass==$password)

to

if($user==$UserName && $pass==$Password)

edited Jul 31 '15 at 07:44

answered Jul 31 '15 at 07:34

Umair Ayub

19,358
14
72
146

STILL MY MAN.. NOT WORKING – deep singh Jul 31 '15 at 07:50
what is output on the page when you submit your form? – Umair Ayub Jul 31 '15 at 07:51
it just remain on the same page.. after submission.. and show the same page – deep singh Jul 31 '15 at 07:52
if u have teamviewer .. plz come on my pc.. and check it – deep singh Jul 31 '15 at 07:53

user1665624 · Accepted Answer · 2015-07-31T08:19:09.047

1

Run the following code ,

   $sql=mysql_query("select* from login where user= '$user' AND pass='$pass'    LIMIT 3 ");
  $data=mysql_fetch_array($sql);

  $UserName= $data['user'];   
  $Password= $data['pass'];
  $type= $data['type'];
  $name= $data['name'];

  if($user==$username && $pass==$password){
  session_start();
  $_SESSION['name']=$name;
  if($type=='admin')
  {
         header("location: index.php");
  }
  else if($type=='vender1')
 {
         header("location: vender1.php");
  }

 }




  else {
        echo 'That information is incorrect, try again <a href="index.php">Click Here</a>';
    exit();
  }
  ?>

You have not used $_POST['log'] in the form ....

In your code you have used if(isset($_POST['log'])) , i have removed that because you have not using 'log' in your form.

edited Jul 31 '15 at 08:19

answered Jul 31 '15 at 08:12

user1665624

91
1
1
7

yes you should start from – user1665624 Jul 31 '15 at 08:38
by your code.. only this part is working else { echo 'That information is incorrect, try again Click Here'; exit(); } – deep singh Jul 31 '15 at 08:38
Check the username and password provided is exactly matching the login table. – user1665624 Jul 31 '15 at 08:42
yea.. its exactly matching my man.... when i give wrong password it takes me to else part .... when i gives right password it remains on the same part. – deep singh Jul 31 '15 at 08:44
if(isset($_POST['log'])) if i do this inplacing of removing the if(isset($_POST['log'])). – deep singh Jul 31 '15 at 08:49
what u say about this @user – deep singh Jul 31 '15 at 08:49
Yes you can do , But try to use **mysqli** Or **PDO** instead of **mysql** – user1665624 Jul 31 '15 at 09:37
ok...... its else part is working.. when password is wrong.. but when password is correct it donot go on index.php and stays on the same page.. can you tell me.. why this is happening man – deep singh Jul 31 '15 at 09:41
@deepsingh Change the line _if($user==$username && $pass==$password)_ to _if($user==$UserName && $pass==$Password)_ – user1665624 Jul 31 '15 at 11:41

multiple level login form not working.. while submitting

2 Answers2