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Edmhar

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Posts posted by Edmhar

    Login & Logout

    in PHP Coding Help

    Posted

     

    Yes you can surround the if/else with HTML code. So long as the file still has a .php extension and you are starting the session before you output the html.

    Example with html

    <?php 
session_start(); //start the session before the html
?>
<html>
<head>
  <title>Page title</title>
</head>
<body>

  <div id="container">
    <div id="header">
      <h1>Header</h1>
      <div id="nav">
        <?php
          // output login or logout links here
          if(isset($_SESSION['logged_in']) && $_SESSION['logged_in'] === true) {
             echo '<a href="logout.php">Logout</a>'; // output logout link
          } else {
             echo '<a href="login.php">Login</a>';   // output login link
          }
        ?>
      </div>
    </div>
    <div id="content">
       page content 
    </div>  
  </div>

</body>
</html>

    Thank you i passed this problem hahaha 

    kindly check my new post simple add and subtract mysql and php

    ADD SUBTRACT PHP MYSQL (affecting database)

    in PHP Coding Help

    Posted

    Hello. im gonna ask how i will make this happen.

    i have  a  mysql database ( StockItem, in, out, availableitem)

    add and subraction only

     

    assuming that  I have value n my StockItem = 50 , in = 0  , out = 0 , availableitem =  ( have a computation of stockitem - in & availableitem + out )

    so i will have form for ( in ) 

    The IN AND OUT are USERINPUT it will have textbox that will input  the value by admin.
    for example i input in FOR (in) textbox = 10 then click the button process it will update the whole table affecting all the field in that row
     

    Login & Logout

    in PHP Coding Help

    Posted

     

    There is no standard built in feature for this.

     

    A simple example would be when a user successfully logs in you set a session variable like this

    $_SESSION['logged_in'] = true; 

    To work out what link to show you check this session variable's value. If it is set to true then you output the logout link. Otherwise you output the login link. Example

    if(isset($_SESSION['logged_in']) && $_SESSION['logged_in'] === true) {
   echo '<a href="logout.php">Logout</a>'; // output logout link
} else {
   echo '<a href="login.php">Login</a>';   // output login link
}

    how about if i put this in html?

    Login & Logout

    in PHP Coding Help

    Posted

     

    There is no standard built in feature for this.

     

    A simple example would be when a user successfully logs in you set a session variable like this

    $_SESSION['logged_in'] = true; 

    To work out what link to show you check this session variable's value. If it is set to true then you output the logout link. Otherwise you output the login link. Example

    if(isset($_SESSION['logged_in']) && $_SESSION['logged_in'] === true) {
   echo '<a href="logout.php">Logout</a>'; // output logout link
} else {
   echo '<a href="login.php">Login</a>';   // output login link
}

    thanks man your great hahaha 

    I love you No homo!  you helping me to finish my Thesis :)

    LOGICAL OPERATION FOR PHP WITH MYSQL (INVENTORY SYSTEM)

    in PHP Coding Help

    Posted · Edited by Edmhar

    the order must not have set value it must be inputed by the admin who in charge in sales.

     

    <?php

include 'connect.php';

//this item order must be input by officer in charge am i correct?


$itemorder = "";

$query ="SELECT * FROM tbl_equipment WHERE Equip_ID ='".$itemid."' AND Equip_name ='".$itemname."' AND EquipVolume = '".$itemstock."' ";
query

echo  $itemname;
echo  $itemid;


echo  $itemstock - $itemorder;

//for example  stock of camera is 15 and a customer rent 5 pcs so in my admin page i will input input the  number of order and  so the other customer will see after they refresh is 10 pcs ony
// It must have effect in my database so im thinking i will use UPDATE but i dont know how to input it in
// then if item went down to 0 it will show The item is 0 with UNAVAILBLE   



?>





    <?php

$conn_err ='Cant Connect';

$mysql_host = 'localhost';
$mysql_itemid = 'root';

$mysql_db = 'mis';

if (!@mysql_connect($mysql_host, $mysql_itemid)||!@mysql_select_db($mysql_db))  {
  die($conn_err);


}

?>

    this is my connect.php

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    Okay,

     

    remove the print_r

    if($_SESSION['login'] && $_SESSION['type'] === 'ADMINISTRATION'){

    You have 3 comparisors ===

    Try with only 2 ==

     

    Other than that its hard to say whats going on. You just have to do some faultfinding within your files and functions.

    you are so good in programming  i hope youre like my professor hahaha :) 

     

    BTW check this link im doing in my inventory system 

    http://forums.phpfreaks.com/topic/283378-logical-operation-for-php-with-mysql-inventory-system/

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

     

    Try this, what do you get ?

    <?php

include '../../core.php';

echo "<pre>";
print_r($_SESSION);
echo "</pre>";

exit();
	
		if($_SESSION['login'] && $_SESSION['type'] === 'ADMINISTRATION'){


?>
<!DOCTYPE html>
<html lang="en">
<head>
	<meta charset="utf-8">
	<link rel="stylesheet" type="text/css" href="../css/main.css"/>
</head>
<body>

This is Admin.
<a href="../../logout.php">Log Out!</a>

</body>


</html>
<?php

}else{
	header('Location:Webpage/index.php');
}

?>

    Array

    (

    [user_id] => 1131

    [type] => ADMINISTRATION

    [login] => 1

    )

     

    This is the output

     

    i think it came from core 

    function loggedin()

    LOGICAL OPERATION FOR PHP WITH MYSQL (INVENTORY SYSTEM)

    in PHP Coding Help

    Posted

    Hi guys i want to ask how to get the value of the in the mysql for  example

    i have a equipments like camera

    so i have  tbl_equipments with field of  EquipmentID, EquipmentName, NoOfEquip,  

     

    for example values

    EquiptmentID : 1001
    EquiptmentName: Camera
    NoOfEquip: 15

    So my customer can see if how many available camera from our website

    EXAMPLE

    A customer rent a camera for 5 pcs 

    if i refresh the page the remaining  camera items is 10 

    and i that customer back that item item

    What maybe the code?

    I have no idea to make this.

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    Hard to follow what you are using at this stage, are you using this ?

    if(!isset($_SESSION['login']) || (isset($_SESSION['type']) && $_SESSION['type'] != 'ADMINISTRATION')) {
	header('Location:Webpage/index.php');
	exit; // stop the script
}

    Its passing you on to Webpage/index.php if session "login" isnt set, and i cannot see its being set anywhere ?

    This means this will always send you to Webpage/index.php. And if Webpage/index.php is checking to see if session "type" is set, it will send you "back" i presume.. and there you have a loop

     

    nope i using even that cause me loop 

     

    I use this

    <?php

include '../../core.php';
	
		if($_SESSION['login'] && $_SESSION['type'] === 'ADMINISTRATION'){


?>
<!DOCTYPE html>
<html lang="en">
<head>
	<meta charset="utf-8">
	<link rel="stylesheet" type="text/css" href="../css/main.css"/>
</head>
<body>

This is Admin.
<a href="../../logout.php">Log Out!</a>

</body>


</html>
<?php

}else{
	header('Location:Webpage/index.php');
}

?>

     

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

     

    I noticed my error on form names but was unable to edit. Rename form inputs to eusername and epassword as commented above.

     

    I dont generally dont provide cut and paste code, only suggested methods. You still have to learn php yourself.

     

    Here is what you need to run the query in my example. Still not tested and it requires a little kung fu from you.

    	if(!empty($err)){
	  echo "<ul><li>".implode("</li><li>",$err)."</li></ul>";
	}
	else{

	  $query = "SELECT LogUsername FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";

	  $query_run = mysql_query($query);

	  if(mysql_num_rows($query_run) == '1'){

	    $row = mysql_fetch_row($query_run);

	    $_SESSION['user_id'] = $row[0];
	    $_SESSION['type'] = $logintype;
	  
            echo "<script>alert('".$logintype." Login')</script>";
	  
            switch($logintype){
	      case 'ADMINISTRATION':
	      header('Location: ../../ADMIN.php');
	      exit();
	      break;
	      default:
	      header('Location: ../../EMPLOYEE.php');
	      exit();
	    }

	  }else{
	    echo "<script>alert('Incorrect Pass or User')</script>";
	  }
        }

    Thank you i got this there is some error but so basic 

    so my problem is now is loop  when i logged in it go loop 

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    You need to change the ID and password form fields to what I suggested and then run the login query

    	else{
	  $query = "SELECT * FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";
	  // run query and set sessions etc

        }

    The code provided by alpine is just an example, it is not fully working code. You need execute the query above for it do anything, then you add your own logic in to redirect the user to correct page based on the users login type (employee or administrator)

    What you think the problem is? hahha

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

     

    Yes, that should work. However I'd change it to this

    <?php

include '../../core.php';

// if the user is not logged in OR they are logged in but they are not part of administration, then redirect to index.php
if(!isset($_SESSION['login']) || (isset($_SESSION['type']) && $_SESSION['type'] != 'ADMINISTRATION')) {
	header('Location:Webpage/index.php');
	exit; // stop the script
}

// load the admin page
?>
<!DOCTYPE html>
<html lang="en">
<head>
	<meta charset="utf-8">
	<link rel="stylesheet" type="text/css" href="../css/main.css"/>
</head>
<body>

This is Admin.
<a href="../../logout.php">Log Out!</a>

</body>
</html>

    I want your this suggestion but it cause Webpage indirect loop 

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    You need to change the ID and password form fields to what I suggested and then run the login query

    	else{
	  $query = "SELECT * FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";
	  // run query and set sessions etc

        }

    The code provided by alpine is just an example, it is not fully working code. You need execute the query above for it do anything, then you add your own logic in to redirect the user to correct page based on the users login type (employee or administrator)

    its already same variable.  but it i change the <? form action= <?php echo $current_file; ?> 

    and it cause access forbidden 

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted · Edited by Edmhar

     

    as a sidenote, you can optimize your loginform with an option list instead of 2 login forms,

    untested version:

    <?php

if(isset($_POST['submit'])){
 
       $err = array();

       $required = array(
	  'eusername',
	  'epassword'
	);
	
	foreach($_POST as $field => $value){
	  if(in_array($field,$required) && empty($value)){
	    $err[] = $field." cannot be empty";
          }
	  else{
	    ${$field} = mysql_real_escape_string($value);
	  }
	}
	
	switch($_POST['id_type']){
	  case 'employee':
	  $logintype = 'EMPLOYEE';
	  break;
	  case 'admin':
          $logintype = 'ADMINISTRATION';
	  break;
	  default:
	  $err[] = "Incorrect login type";
	}
	
	if(!empty($err)){
	  echo "<ul><li>".implode("</li><li>",$err)."</li></ul>";
	}
	else{
	  $query = "SELECT * FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";
	  // run query and set sessions etc

        }
	
}

?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">

ID: <input type="text" name="username"> </br>

Password: <input type="password" name="password">

Type:
<select name="id_type">
<option value="employee">Employee</option>
<option value="admin">Admin</option>
</select>

<input type="submit" name="submit" id="adminsubmit" value="Log in">

</form>

    Hmm.. Hi I try your code but when i run it will go something like error. webpage do not display like that but it will to my adminhome but i will have no output. 

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

     

    as a sidenote, you can optimize your loginform with an option list instead of 2 login forms,

    untested version:

    <?php

if(isset($_POST['submit'])){
 
       $err = array();

       $required = array(
	  'eusername',
	  'epassword'
	);
	
	foreach($_POST as $field => $value){
	  if(in_array($field,$required) && empty($value)){
	    $err[] = $field." cannot be empty";
          }
	  else{
	    ${$field} = mysql_real_escape_string($value);
	  }
	}
	
	switch($_POST['id_type']){
	  case 'employee':
	  $logintype = 'EMPLOYEE';
	  break;
	  case 'admin':
          $logintype = 'ADMINISTRATION';
	  break;
	  default:
	  $err[] = "Incorrect login type";
	}
	
	if(!empty($err)){
	  echo "<ul><li>".implode("</li><li>",$err)."</li></ul>";
	}
	else{
	  $query = "SELECT * FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";
	  // run query and set sessions etc

        }
	
}

?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">

ID: <input type="text" name="username"> </br>

Password: <input type="password" name="password">

Type:
<select name="id_type">
<option value="employee">Employee</option>
<option value="admin">Admin</option>
</select>

<input type="submit" name="submit" id="adminsubmit" value="Log in">

</form>

    Hey man , it cause an  error 

     

    Notice: Undefined variable: eusername in C:\xampp\htdocs\MIS\login\loginform.php on line 36

     

    Notice: Undefined variable: epassword in C:\xampp\htdocs\MIS\login\loginform.php on line 36 

     

    Thank you :))

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

     

    as a sidenote, you can optimize your loginform with an option list instead of 2 login forms,

    untested version:

    <?php

if(isset($_POST['submit'])){
 
       $err = array();

       $required = array(
	  'eusername',
	  'epassword'
	);
	
	foreach($_POST as $field => $value){
	  if(in_array($field,$required) && empty($value)){
	    $err[] = $field." cannot be empty";
          }
	  else{
	    ${$field} = mysql_real_escape_string($value);
	  }
	}
	
	switch($_POST['id_type']){
	  case 'employee':
	  $logintype = 'EMPLOYEE';
	  break;
	  case 'admin':
          $logintype = 'ADMINISTRATION';
	  break;
	  default:
	  $err[] = "Incorrect login type";
	}
	
	if(!empty($err)){
	  echo "<ul><li>".implode("</li><li>",$err)."</li></ul>";
	}
	else{
	  $query = "SELECT * FROM tbl_account WHERE LogUsername='".$eusername."' AND LogPassword = '".$epassword."' AND type = '".$logintype."'";
	  // run query and set sessions etc

        }
	
}

?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">

ID: <input type="text" name="username"> </br>

Password: <input type="password" name="password">

Type:
<select name="id_type">
<option value="employee">Employee</option>
<option value="admin">Admin</option>
</select>

<input type="submit" name="submit" id="adminsubmit" value="Log in">

</form>

    HEY THANK YOU :))) VERY USEFUL THANKS :)

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    every page must check if the user that is accessing it has the appropriate permission to do so.

     

    since you have a type value in your database table, you would test if the type of the current user is an admin or an employee.

    like this? this is the home of my admin i put this

     

    <?php

include '../../core.php';
	
		if($_SESSION['login'] && $_SESSION['type'] === 'ADMINISTRATION'){


?>
<!DOCTYPE html>
<html lang="en">
<head>
	<meta charset="utf-8">
	<link rel="stylesheet" type="text/css" href="../css/main.css"/>
</head>
<body>

This is Admin.
<a href="../../logout.php">Log Out!</a>

</body>


</html>
<?php

}else{
	header('Location:Webpage/index.php');
}

?>

    2 Types of Users problem (Problem: Dont have Limitation)

    in PHP Coding Help

    Posted

    HI, Guys Im back! After i successfully found my solution to problem earlier. I want to ask what code i will add to have limitation to employee to admin can access, 

    My problem is when My Employee Logged in he will direct to localhost/MIS/Webpage/Employee/home.php
    This is the correct for my employee

     

    120m6gg.jpg

     

    but when i changed the address to localhost/MIS/Webpage/Admin/home.php

    My Employee can access the admin homepage. this is the problem i want to have limitation of  my employee access.

     

    2ex4s28.jpg

     

    so this is my codes of my index.php

    <?php
require 'core.php';
require 'connect.php';


if (loggedin()) {
	 if($_SESSION['type'] == 'ADMINISTRATION'){
		header('Location:../Mis/Webpage/Employee/home.php');
	 }else if($_SESSION['type'] == 'EMPLOYEE'){
		 header('Location:../Mis/Webpage/Admin/home.php');
	 }
}
else{
	header('Location:Webpage/index.php');
}



?>

    this is my loginform 

    <?php
include '../../Mis/connect.php';
include '../../Mis/core.php';

if(isset($_POST['eusername']) && isset($_POST['epassword'])){
	if(!empty($_POST['eusername']) && !empty($_POST['epassword'])){
		$user = mysql_real_escape_string($_POST['eusername']);
		$pass = mysql_real_escape_string(md5($_POST['epassword']));
			$query = "SELECT * FROM tbl_account WHERE LogUsername='".$user."' AND LogPassword = '".$pass."' AND type = 'EMPLOYEE'";
				if($query_run = mysql_query($query)){
					$query_num_rows = mysql_num_rows($query_run);
						if($query_num_rows == 0){
							echo "<script>alert('Incorrect Pass or User')</script>";
						}else{
							 $user_id = mysql_result($query_run, 0, 'LogUsername');
							 $_SESSION['user_id']=$user_id;
							 $_SESSION['type'] = "EMPLOYEE";
							 echo "<script>alert('Employee Login')</script>";
							 header('Location: ../../Mis/index.php');
						}

				}else{
					echo "<script>alert('Connecting Failed')</script>";
				}


	}else{
		echo "<script>alert('Sorry, You must supply Username/Password...')</script>";

	}
}

if(isset($_POST['username']) && isset($_POST['password'])){
	if(!empty($_POST['username']) && !empty($_POST['password'])){
		$user = mysql_real_escape_string($_POST['username']);
		$pass = mysql_real_escape_string(md5($_POST['password']));
			$query = "SELECT * FROM tbl_account WHERE LogUsername='".$user."' AND LogPassword = '".$pass."' AND type = 'ADMINISTRATION'";
				if($query_run = mysql_query($query)){
					$query_num_rows = mysql_num_rows($query_run);
						if($query_num_rows == 0){
							echo "<script>alert('Incorrect Pass or User')</script>";
						}else{
							 $user_id = mysql_result($query_run, 0, 'LogUsername');
							 $_SESSION['user_id']=$user_id;
							 $_SESSION['type'] = "ADMINISTRATION";
							 echo "<script>alert('Admin Login')</script>";
							 header('Location: ../../Mis/index.php');
						}
				}else{
					echo "<script>alert('Connecting Failed')</script>";
				}


	}else{
		echo "<script>alert('Sorry, You must supply Username/Password...')</script>";

	}
}

?>
<div id="employee">

<form action="<?php echo $current_file; ?>" method="POST">
Employee ID: <input type="text" name="eusername"> </br>
Password: <input type="password" name="epassword">
<input type="submit" id="employeesubmit" value="Log in">
</form>
</div>



<div id="admin">

 <form action="<?php echo $current_file; ?>" method="POST">
Admin ID: <input type="text" name="username"> </br>
Password: <input type="password" name="password">
<input type="submit" id="adminsubmit" value="Log in">
</form>
</div>

    This is my core.php

    <?php
ob_start();
session_start();
$current_file = $_SERVER['SCRIPT_NAME'];


function loggedin() {
  if (isset($_SESSION['user_id'])&&!empty($_SESSION['user_id'])) {
    return true;
  } else {
    return false;;
  }

}


function adminloggedin() {
  if (isset($_SESSION['user_id'])&&!empty($_SESSION['user_id'])) {
    return true;
  } else {
    return false;;
  }

}



?>
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