PHP - Multi Level Menu Help

I want to create an array from a file with this structure

TITLE1, HEADER1, key11, key12, key13, key14
TITLE2, HEADER2, key21, key22, key23, key24
.
.
.

How I can get a foreach with these elements:

$title="TITLE"
$header="HEADER"
$key="(one random key)"

If needed, I can change the file structure too.

I want to have a multi level dropdown selection tool come out of an array (not database; almost all of available codes on the internet are based on mysql databse). I have an array as each element has this format: "Country | City | ID". Using explode, I want to make a dropdown to choose the country, then choosing the cities of the chosen country. Finally, in a simple search box make action="result.php" to lead to result.php?q=$id

Thanks for your kind attention 

Mmm i'm not sure if i'm approaching this correctly.

Anyways, I wanted to create a multi-level thread, Where the child thread would be after the parent thread.

Something like this:

Hello World
       - Hello World
               - Hi!
               - What's up?
       - Hi Bob!
Goodbye World
       - ....This guy got issues
       - Dude! You should watch Despicable Me! "It's so fuzzy i'm going to dieee!"


Anyways, so my table looks something along the line of this:

|    post id   |    parent_id    |            title            |      body    |   date      |
      1                |       NULL        |    Hello                  | Hi, i'm K    |   date
      2                |       NULL        |    Thread One       | Body One  |   date
      3                |          1           |    Hello World | Body One  |   date
      4                |          2           |    Thread post 4   | Body One  |   date
      5                |          1           |    Thread post 5   | Body One  |   date
      6                |          1           |    Thread post 6   | Body One  |   date
      7                |          3           |    Hi What's up    | Body One  |   date


And here's the PHP code, although it's not very neat I want to clean it up and make it neater.  I can only go two level deep so far.   Any suggestion or recommendations? 

Using while & for:


// Select * Threads
$q = "SELECT * FROM test_db ORDER BY post_id ASC";
$r = @mysql_query($q);



while ($pid_row = @mysql_fetch_object($r)) {

// Array
$pid_arr[] = $pid_row->post_id; // Assign Indexes to Post ID
$ppid_arr[] = $pid_row->parent_id; // Assign Indexes to ParentID
$md_arr = array($pid_arr, $ppid_arr);  // Create multi-dimensional array matrix

$parent_id = $pid_row->parent_id;




}

for ($i = 0; $i < count($pid_arr); $i++) {

for ($a = 0; $a < count($ppid_arr); $a++) {
$pid = $pid_arr[$i]; // pid = post id
$ppid = $ppid_arr[$i]; // ppid = parent id

//echo "<div>~~ PID $pid ~~ PPID $ppid ~~</div>";

if ($ppid === NULL) {
$ppid_is_null = " & parent_id is NULL";
} else {
$ppid_is_null = NULL;
}

// Select Top level Thread
$z = "SELECT * FROM test_db WHERE post_id = '$pid'";
$y = @mysql_query($z);

while ($t_row = @mysql_fetch_object($y)) 
{

$title = $t_row->title;  // title of child thread
$level = $t_row->level;  // level of child thread...is this useful?
}




// Display top level thread lvl 1
if ($ppid === NULL) {  // If the parent_id is NULL, display top level thread
echo '<div>PID: ' . $pid . ' || PPID: ' . $ppid . " || Title: " . $title . "</div><br>\n";

$post = $pid; // PostID = 1
$parent = $ppid; // Parent ID ex: [1]

// Get child id, where parent id = [1]
// trying to get the child thread here...but it's not working
$child_q = "SELECT * FROM test_db WHERE parent_id = '$post'";
$child_r = @mysql_query($child_q);

while($c_row = @mysql_fetch_object($child_r)) {
$c_arr[] = $c_row->post_id;
$c_ppid_arr[] = $c_row->parent_id;
}
for ($n = 0; $n < count($c_arr); $n++) {
if ($c_ppid_arr[$n] === $post) {
echo "<div>$c_arr[$n]</div>";



}

}
break;

//echo "<div>PID: $pid</div>";




} 




break;
}

}



Using While:

<?php 
require_once('db_connection');
echo '<hr><hr><br>';

// I just realized that selecting * where parent id is null might limit the amount of rows retrieved.

// Query to select threads
$n = NULL;
$q = "SELECT * FROM test_db WHERE (test_db.parent_id is NULL) Order By date ASC";
$r = @mysql_query($q);

while ($threads = @mysql_fetch_object($r)){
$pid = $threads->post_id;
$sid = $threads->subject_id;
$parent_id = $threads->parent_id;
$title = $threads->title;
$body = $threads->body;



echo '<div>PostID: ' . $pid . ' || Parent: ' . $parent_id . ' || SubjectID: ' . $sid . ' || Title: ' . $title . ' || Body: ' . $body . '</div>';

$q2 = "SELECT * FROM test_db WHERE parent_id = '$pid' Order by date DESC";
$r2 = @mysql_query($q2);

while ($level2 = @mysql_fetch_object($r2)) {
$level2_pid = $level2->post_id;
$level = 1.5 *($level2->level); // lvl 2: 3em


echo '<div style="text-indent:' . $level . 'em">' . 'PostID: ' . $level2->post_id . ' || ParentID: ' . $level2->parent_id . '</div>';


$q3 = "SELECT * FROM test_db WHERE parent_id = '$level2_pid'";
$r3 = @mysql_query($r3);
while ($level3 = @mysql_fetch_object($r3)) {

$level3 = 1.5 + $level;

echo '<div style="text-indent:' . $level3 . 'em">' . 'PostID: ' . $level3->post_id . ' || ParentID: ' . $level3->parent_id . '</div>';

}






}


}


?>


I'm not a comp. sci major or anything, so I'm learning this as I go.  Any help would be greatly appreciated.   It seems like the while & for is more practical, but i'm not quiet sure.  What am I doing wrong?

- K

I just joined this forum today ^_^  Hope to be good friends with you guys and gals

Hi -
I've got a program to allow the user to upload a file and then select, for each column
of an existing table, the column from the new source table to insert.  However, what
I need to do is do the insert using the actual names, but show the user a drop-down
menu of the LABELS they've chosen for each of these real, target columns.  Is there
a "clean" or "tricky" way of doing this?  The code to do the inserts works by just reading
the translation table that has two columns --  source column fieldname and target column
(in the existing table) fieldnames (both real fieldnames, of course).

Here's what I've got:

<form action="index.php" method="post">
   <input type="hidden" id="unique" name="unique" value="<?php echo $unique_nm; ?>" />
   <input type="hidden" id="savetotable" name="savetotable" value="1" />
   <?php foreach($LabelList as $c_name){ ?>
      <div>
         <div style="float: left; width: 300px;"; >
            <?php echo $c_name; ?>
         </div>
         <div style="float: left; width: 400px;">
            <select id="<?php echo $c_name; ?>" name="<?php echo $c_name; ?>">
               <option value=""></option>
               <?php foreach($SourceColumns as $h_name){ ?>
                  <option value="<?php echo trim($h_name); ?>">
                     <?php echo trim($h_name); ?>
                  </option>
               <?php } ?>
            </select>
         </div>
         <div style="clear: both;"></div>
      </div>
   <?php } ?>
   <input type="submit" value="Save to Database" />
</form>

So, while I could load the data into the targetfile pretty easily using the
real target table column names, how can I present the user with a drop-down
menu of LABELS, but insert the data into the REAL target fields?

Here's the list of ACTUAL fields:
   $output = array();
   $gather = 'select FieldName from MapTable where User = UserID'
   $this->call($gather);
      foreach($this->data as $row){
         $ActualFields[] = $row['COLUMN_NAME'];
      }
      return $ActualFields;
   }

Here's the list of LABELS (phony field names):
   $output = array();
   $gather = 'select LabelName from MapTable where User = UserID'
   $this->call($gather);
      foreach($this->data as $row){
         $LabelList[] = $row['COLUMN_NAME'];
      }
      return $LabelList;
   }

The list of fields from the sourcefile the user is loading
are all actual, real names ($SourceColumns)

Hi everyone,
I'm very new to PHP and have been following Kevin Yank's "How to build a db driven website using php & mysql" 3rd ed. and am working through Chapter 6, where a basic search page is created. What I would like to be able to do is, in the results, display not only the joketext (as done in the book), but also the author's name (which is located in a different table in the db), as well as the date it was submitted (in the same db as joketext), possibly a link to another site and extra info like the joke category. In general, I would like to be able to understand how to adjust the php code to add data related to the results of any given search. The information is already related through the database, how do I use it in the results of a search?

I've used the same names of variables and tables as used in the book.

Thank you in advance for any help you can offer! I really appreciate it.

Below is the code for the part of the code I would like to change (highlighted parts) followed by the code for whole page (2 main sections "search" and "results". In the main code, the excerpt below is located a few lines from the bottom (in "results"):

Note: the only variable currently displaying in the results in $joketext (also highlighted). Ok, enough explaining!

/////////// EXCERPT : PART I WANT TO DISPLAY DIFFERENTLY //////////////

<?php
echo "<li id=\"jump\">
<article class=\"entry\">
<header>
<h3 class=\"entry-title\"><a href=''$VAR for email or weblink from author table">$VAR for author name from author table</a></h3>
</header>
<div class=\"entry-content\">
<p>$joketext</p></div>
<footer class=\"entry-info\">
<abbr class=\"published\" title=\"2011-09-22T14:07:00-07:00\">$VAR for date uploaded in joke table</abbr>
<p>$VAR for joke category from jokecategory table</p>
</footer>
</article>
</li>";
}
?>
//////////////////////////// MAIN CODE ///////////////////////

<html>
<body>
.
<header></header>
.
<section id="search">

<?php

$dbcnx = @mysql_connect('localhost', 'root', 'password');

if (!$dbcnx) {
exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}

if (!@mysql_select_db('ijdb')) {
exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
}

$authors = @mysql_query('SELECT id, name FROM author');
if (!$authors) {
exit('<p>Unable to obtain author list from the database.</p>');
}

$cats = @mysql_query('SELECT id, name FROM category');
if (!$cats) {
exit( '<p>Unable to obtain category list from the database.</p>');
}

$themes = @mysql_query('SELECT id, name FROM theme');
if (!$themes) {
exit( '<p>Unable to obtain category list from the database.</p>');
}

$geofoci = @mysql_query('SELECT id, name FROM geofocus');
if (!$geofoci) {
exit( '<p>Unable to obtain category list from the database.</p>');
}

?>

<form class="searchField" name="input" action="main_search.php#jump" method="post">
<input type="text" name="searchtext">
<input type="submit" value="Search">
<ul>
<li>
<label><select name="aid" size="1">
<option selected value="">Any Author</option>
<?php
while ($author = mysql_fetch_array($authors)) {
$aid = $author['id'];
$aname = htmlspecialchars($author['name']);
echo "<option value='$aid'>$aname</option>\n";
}
?>
</select></label>
</li>
<li>
<label><select name="cid" size="1">
<option selected value="">Any Category</option>
<?php
while ($cat = mysql_fetch_array($cats)) {
$cid = $cat['id'];
$cname = htmlspecialchars($cat['name']);
echo "<option value='$cid'>$cname</option>\n";
}
?>
</select></label>
</li>
<li>
<label><select name="tid" size="1">
<option selected value="">Any Theme</option>
<?php
while ($theme = mysql_fetch_array($themes)) {
$tid = $theme['id'];
$tname = htmlspecialchars($theme['name']);
echo "<option value='$tid'>$tname</option>\n";
}
?>
</select></label>
</li>
<li>
<label><select name="gfid" size="1">
<option selected value="">Any Region</option>
<?php
while ($geofocus = mysql_fetch_array($geofoci)) {
$gfid = $geofocus['id'];
$gfname = htmlspecialchars($geofocus['name']);
echo "<option value='$gfid'>$gfname</option>\n";
}
?>
</select></label>
</li>
<li><a href="">Closing Date</a></li>
</ul>
</form>
</section>

<section id="results">
<?php

$dbcnx = @mysql_connect('localhost', 'root', 'password');

if (!$dbcnx) {
exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}

if (!@mysql_select_db('ijdb')) {
exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
}

// The basic SELECT statement

$select = 'SELECT DISTINCT id, joketext';
$from   = ' FROM joke';
$where = ' WHERE 1=1';

$aid = $_POST['aid'];
if ($aid != '') { // An author is selected
$where .= " AND authorid='$aid'";
}

$cid = $_POST['cid'];
if ($cid != '') { // A category is selected
$from .= ', jokecategory';
$where .= " AND joke.id=jokecategory.jokeid AND categoryid='$cid'";
}

$tid = $_POST['tid'];
if ($tid != '') { // A theme is selected
$from .= ', joketheme';
$where .= " AND joke.id=joketheme.jokeid AND themeid='$tid'";
}

$gfid = $_POST['gfid'];
if ($gfid != '') { // A region is selected
$from .= ', jokegeofocus';
$where .= " AND joke.id=jokegeofocus.jokeid AND geofocusid='$gfid'";
}

$searchtext = $_POST['searchtext'];
if ($searchtext != '') { // Some search text was specified
$where .= " AND joketext LIKE '%$searchtext%'";
}

?>

<ol id="results-list">

<?php
$jokes = @mysql_query($select . $from . $where);
if (!$jokes) {
echo '</table>'; exit('<p>Error retrieving jokes from database!<br />'.
'Error: ' . mysql_error() . '</p>');
}

while ($joke = mysql_fetch_array($jokes)) {
$id = $joke['id'];
$joketext = htmlspecialchars($joke['joketext']);
echo "<li id=\"jump\">
<article class=\"entry\">
<header>
<h3 class=\"entry-title\"><a href=''>variable title</a></h3>
</header>
<div class=\"entry-content\">
<p>$joketext</p></div>
<footer class=\"entry-info\">
<abbr class=\"published\" title=\"2011-09-22T14:07:00-07:00\">Sept. 22, 2011</abbr>
</footer>
</article>
</li>";
}

?>

</ol>
</section>
.
<footer></footer>
.
.
</body>
</html>

When I use ini_get to check my error_reporting level, I get a weird value ( 4983 ) that I can't find anywhere by googling.

Also not on the list of possible levels here ( http://itech.hubpages.com/hub/php-error_reporting ) .. does anyone know how to determine what exactly this level is?

Thanks!

Hi All,

I have a function that creates an array using a few sql quereys.  I then use the array to output to html.  I am wanting to add more data to the array to populate another column in the created table.

I am unsure how to add $miqty to the array and then output it in the Qty On Order column. 

I feel like i need to add the [] to

$menuItemsInCat[$mcatname][$miid] = $miname;

section but i am not sure where in there it should be.

 

<?php if (session_status() == PHP_SESSION_NONE) {
session_start();
}
if (!isset($_SESSION['user_id'])){
	header("location: index.php");
	exit;
}
//name of the pagea
$_SESSION['this_page'] = 'new-menu';

function getMenuItems($conn){
	$output = '';
	$stmt = $conn -> query("
	                         SELECT menu_category_name
	                         FROM ssm_menu_items
	                         INNER JOIN  ssm_menu_category on menu_item_category_id = menu_category_id
	                         ORDER BY menu_category_display_order
	                         ");
	//create an empty array of all of the menu categories that are in use
	foreach ($stmt as $item){
		$menuItemsInCat[$item['menu_category_name']] = [];
	}
	$stmt = $conn -> prepare("
	                         	SELECT menu_item_id, menu_item_name, menu_category_name, sum(menu_item_qty)
								FROM ssm_menu_items mi
								INNER JOIN ssm_menu_category mcat ON mi.menu_item_category_id = mcat.menu_category_id
								left join ssm_menu_order USING (menu_item_id)
								GROUP BY menu_item_id
	                         ");
	$stmt -> execute();
	$stmt -> bind_result($miid, $miname, $mcatname, $miqty);
	while ($row = $stmt -> fetch()) {
		//put items into the blank array created above under their correct category
		
		$menuItemsInCat[$mcatname][$miid] = $miname;
	}

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

		//foreach thing in $menuItemInCat array there is $menucat array associated with $menuit(ems) array
		//we want the menu cat
	foreach ($menuItemsInCat as $menucat => $menuit) {
		$output .= "<tbody>";
		$output .= "<tr class='bg-secondary text-white text-center'><th>$menucat</th>";
		$output .= "<th>Qty On Order</th><th>Manage</th></tr>";
		//foreach thing in menu items array there is an array of ids and an array of items
		foreach ($menuit as $itemId => $itemName) {
			$output .= "<tr><td>$itemName</td>";
			$output .= "<td></td>";
			$output .= "<td><div class='btn btn-primary'>Manage</div></td></tr>";
		}
		$output .= "</tbody>";
	}
	return $output;
}



?>
<?php include '_includes/head.php'; ?>
<div class="container-fluid">
	<div class="row">
		<?php include '_includes/header.php'; ?>
	</div>
	<div class="row" >
		<div class="col-sm-2 p-0 bg-dark text-light">
			<?php include '_includes/nav.php'; ?>
		</div>
		<div class="col-sm-10" style="height: calc(100vh - 80px);overflow:scroll;">
			<div class="mt-3">
				<table class="table table-striped table-hover table-bordered text-center align-middle">
					<?= getMenuItems($conn) ?>
				</table>
			</div>
			<div class="col-sm-12"><?php include '_includes/footer.php'; ?></div>
		</div>
	</div>
	
</div>
<script>
	//set sidebar active indicator
	//XX = name of parent if in dropdown eg "sheet"
	if(document.getElementById('menu')){
		document.getElementById('menu').classList.add('show')
	}
	//nav button ID
	if(document.getElementById('newMenu')){
		document.getElementById('newMenu').classList.add('blOrange')
	}
</script>

As always your help is very appreciated.

Edited February 5, 2020 by Adamhumbug

Hi, I would like to make a login page with 2 different user level, Admin & Staff.  How can i do it using d code below.

Thank u.



<?php
$host="localhost";
$username="root"; 
$password=""; 
$db_name="profile";
$tbl_name="company";


mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("Cannot Select Database");

// username and password sent from form
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];


$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);


if($count==1){

session_register("myusername");
session_register("mypassword");
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
?>

i try to create login page with two different user level for example admin and staff. i did not get any error on my code but it just did not direct to the page it been set to. it just display wrong username or password. i not really sure what is wrong.

here is the code

loginForm.php

Code: [Select]
          <form action="login.php" method ="post">

          <table>
          <tr><td>Usernama</td>
              <td><input name="username" type="text" size = "15" maxlength = "15"/></td></tr>
     
      <tr><td>Password</td>
              <td><input name="password" type="password" size = "15" maxlength = "15"/></td></tr>
      </table>
 
          <br><input name="submit" type ="submit" value ="Login"/></td>
 
          </form>

login.php

Code: [Select]
<?php
ob_start();
$host="localhost"; 
$user="root"; 
$pass=""; 
$db_name="office"; 
$tbl_name="login"; 

mysql_connect("$host", "$user", "$pass")or die("cannot connect"); 
mysql_select_db("$db_name")or die("Cannot Select Database"); 

// username and password sent from form 
$sername=$_POST['username']; 
$password=$_POST['password']; 

$username = stripslashes($username); 
$password = stripslashes($password); 
$username = mysql_real_escape_string($username); 
$password = mysql_real_escape_string($password); 

$sql="SELECT * FROM daftarPenyelia WHERE user='$username' AND pass='$password' AND userLevel='$userLevel'"; 
$result=mysql_query($sql); 

// Mysql_num_row is counting table row 
$count=mysql_num_rows($result); 

if ($count == $userLevel)
{
   if ($userLevel == 1) 
  { 
        $_SESSION['username'] = $username; 
        $_SESSION['password'] = $password; 
        header("location:adminMenu.php"); 
   } 
   else if ($userLevel == 2)
  { 
        $_SESSION['username'] = $username; 
        $_SESSION['password'] = $password; 
        header("location:staffMenu.php"); 
   } 
 }
else
{ 
echo "Wrong Username or Password"; 
} 
?>

can someone help me with this code and tell me what is wrong so that i can fix them

Hello everybody,

i am working on n level category tree structure, for your information the category tree structure image is attached with this post.

Let's consider, each of the category have hundreds( n level ) of category in each, i want to traverse each category without using recursive functions, reason being recursive functions are very slow,  is there any way to do this??

Thanks,
phpeid

The way I understand posting a form to work is that when you hit submit the information is attached to the end of the url of the new page where your script can then get it and use it.

I have a form that is an include on a page, like so:

thepage.php
include: theform.php

When i fill out the form and hit submit.. the next page acts as if it has not received any of the posted information.

example: echo $_POST['input1'] results in nothing

If i go directly to theform.php and fill out and submit the form it works fine.

example: echo $_POST['input1'] results in whatever was typed in the first box.

So how can I get it to work while being included?

Hello.  Basically I have created a form that you fill in and an email gets sent to your in box and you have an access_level of 1.  Within this email there is one field.  A validation_id.  Now what im trying to do is when you click this link it opens the webpage and updates your access_level to 2.  I have wrote my query and array and i can see exactly which bit is wrong.

Its where i have

validation_id= $validation_id")

Now its not working and I can see the error.  Problem is im too new at this to understand why its wrong and how to fix it can anyone help?  Here is the rest of the code.  Also could you please explain?  I would really like to understand it before i move any further.
p.s.  I have already selected my database in the included file

      <? include('database_name');
session_start();


  $validation_id =$_GET['validation_id'];
  $FullRec__query=sprintf("SELECT * FROM Members WHERE validation_id= $validation_id"); 

  $FullRec = mysql_query($FullRec__query, $database name) or die(mysql_error());

  $FullRecArray = mysql_fetch_array($FullRec);

  $UdateAccessQuery = sprintf("UPDATE Members SET access_level = '2' WHERE validation_id = $validation_id");

  mysql_query($UdateAccessQuery, $database_name) or die(mysql_error());
 ?>

i need to know how i can check a users level in my login.php page it works i have 

$sql = "SELECT * FROM $tbl_name WHERE username = '$username' AND password='$password'";
	$result = mysql_query($sql);
	$count = mysql_num_rows($result);
	$row = mysql_fetch_assoc($result);
	$user_level = $row['user_level'];

	if($count == 1) {

		$_SESSION['loggedIn'] = true;
		session_write_close();
		header("Location: index.php");

	} else {

		echo "The username or password you entered is incorrect!";
	} if($row['user_level'] == 1) {

		header("Location: admin.php");
		
	} else if($row['user_level'] == -1) {

		header("Location: banned.php");
	}

but i need to know how to check it in another file because it is not working i am trying to add it to admin.php to check the users level & if they are not admin then echo you are not admin.  <-- it says that although the user is an administrator it is saying they are not.   This is what i have in admin.php

<?php

require 'connect.php';

session_start();

$sql = "SELECT * FROM $tbl_name WHERE username = '$username' AND password='$password'";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$user_level = $row['user_level'];

if(!isset($_SESSION['loggedIn'])) {

	echo "You are not currently logged in and to view this page you must be logged in to have access. <a href='login.php'> You can login here </a>";
	die();

} if($row['user_level'] == 1) {

	//DO NOTHING
} else {

	echo "Your not an administrator so you are denied access to this page.";
	die();
}

?>

Edited by Tom8001, 24 November 2014 - 03:02 PM.

CREATE TABLE user (username varchar(20),password varchar(20),level varchar(20),PRIMARY KEY(username));

INSERT INTO `user` VALUES ('a', 'pass1', 'admin');
INSERT INTO `user` VALUES ('b', 'pass2', 'admin');
INSERT INTO `user` VALUES ('c', 'pass3', 'user');


This is my database

and

Registration.php

Code: [Select]
<html>
<head>
<script type="text/javascript">
  function a()
{
   var x = document.register.username.value;
   var y = document.register.pass.value;
   var z = document.register.pass2.value;
   
   if(x==""&& y==""&& z=="")
   {
    alert("Please insert all message!");
    return false;
   }
   if(x=="")
   {
     alert("Please insert an username!");
     return false;
   }
   if(y=="")
   {
     alert("Please insert an password!");
     return false;
   }
   if(z=="")
   {
     alert("Please insert an password2!");
     return false;
   }
   
   if (y!=z)
   {
     alert("Your passwords did not match");
     return false;
   }
   



}
</script>
</head>
<?php
    mysql_connect("localhost","root") or die(mysql_error());
    mysql_select_db("cute") or die(mysql_error());


     if (isset($_POST["sub"]))
     {
        $_POST['pass'] = md5($_POST['pass']);
      if (!get_magic_quotes_gpc())
       {
        $_POST['username'] = addslashes($_POST['username']);
        $_POST['pass'] = addslashes($_POST['pass']);
      }


       $usercheck = $_POST["username"];

       $check = mysql_query("SELECT username FROM regis WHERE username = '$usercheck'") or die(mysql_error());

       $check2 = mysql_num_rows($check);
        //if the name exists it gives an error
     if ($check2 != 0)
     {
      echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('Sorry, the username" ." ".$usercheck." ". "is already in use.')</SCRIPT>");
      echo ("<SCRIPT LANGUAGE='JavaScript'>setTimeOut(window.location = 'registration.php',1)</script>");

     }

      else if($_POST['username'] && $_POST['pass'] && $_POST['pass2'] )
     {
      $insert = "INSERT INTO regis(username, password) VALUES ('".$_POST['username']."', '".$_POST['pass']."')";
      $add_member = mysql_query($insert);
      echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('Registration had been succesfully added :)')</SCRIPT>");
      echo "<meta http-equiv='refresh' content='0; url=login.php'>";
     }

     }
     
?>
<body>



  <form name="register" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"  onsubmit="return a()">
  <table border='0'>
  <tr><td>Username:</td><td><input type="text"name="username" maxlength="60"></td></tr>
  <tr><td>Password:</td><td><input type="password" name="pass" maxlength="10"></td></tr>
  <tr><td>Confirm Password:</td><td><input type="password" name="pass2" maxlength="10"></td></tr>
  <tr><th colspan=2><input type="submit" name="sub" value="Register"></th></tr></table>
  </form>
 

  </body>
 </html>

My main problem when in registration iam did not put user level field because it is not secure but how to manage or how to detect when someone registering he or she is user or admin?

Code: [Select]
$sqlCommand = "CREATE TABLE products (
id int(11) NOT NULL auto_increment,
product_name varchar(255) NOT NULL,
price varchar(16) NOT NULL,
details text NOT NULL,
category varchar(16) NOT NULL,
subcategory varchar(16) NOT NULL,
m_tag text NOT NULL,
date_added date NOT NULL,
PRIMARY KEY (id),
UNIQUE KEY product_name (product_name)
) ";.
this is currently my database, i would like to create a stock level count so im assuming i need a new field called (stock)

also on my homepage i would like a list of best selling items so im probably going to need a (total sold)

my website has a basket, when you click check out it goes to paypal so you can pay for your transaction. i have a ipn script to verify with payal.

but i dont have a clue how to update my stock level after each transaction. can someone please help and point me in the right direction