PHP - Difficulty Of Creating Online Forum

Hello everyone. Before I start, I'm relivily new to PHP but have been doing it for a few weeks or so, my latest problem that i've been trying to crack for a few days now is using the "imagecopy" command in php, I'm working on an avatar script that lets users design there own avatar, selecting clothes, hair, eyes, etc. I've only just started and i've ran into a rather large problem so don't seem to be able to move any further,

This is the code i have

Quote

<?php
ob_start ();
/* background */
$bg = imagecreatefrompng('bg.png');
/* foreground */
$fg = imagecreatefrompng('fg.png');
/* frontground */
$fgg = imagecreatefrompng('fgg.png');


$x = imagesx($bg);
$y = imagesy($bg);


imagecopy($bg, $fg, $fgg, $x, $y);

header('Content-Type: image/png');
imagepng($bg);

imagedestroy($bg);
imagedestroy($fg);
imagedestroy($fgg);
?>


However for some reason, probably a simple fault, it's just giving me a broken image link? I'm not sure why it's doing this if you want to see exactly what it's doing please look at http://chat.blastgames.org/index.php Like i say i'm clueless so any help would be much appreciated,

Also if anyone knows of any downloadable php avatar scripts that lets you customize things like hair, clothes etc. please tell me, I've been searching for one for a while,

Thanks for reading , Any help will be much appreciated.

Hi, I have a form that I want to allow users to be able to add specific links, some are set types (domains) and just require the iser to enter the link, however, one is for whatever they wish and they must enter a name to describe it.

Now im going to enter it into a database with the following columns:

USER_ID || LINK_URL || LINK_TYPE || LINK_NAME

Link name is optional (only needed if the link type is their own custom one), the link types are ID's related to a table of link types (facebook, twitter etc).

Now I'm assuming that to differentiate the form elements for insertion into the database I would need to have a hidden form element with the type ID inside? However I don't really know how to go on from here.


A small version of the form would be

<input type="text" name="facebook_link" />
<input type="hidden" name="facebook_id value="1" />

<input type="text" name="twitter_link" />
<input type="hidden" name="twitter_id value="2" />

<input type="text" name="own_site" />
<input type="text" name="own_site_description" />
<input type="hidden name="own_site_id" value="3" />


However, I'm assuming this is wrong because I'll need to enter everything into an array won't I? I just can't get my head round it to out put thi results into a foreach statement.


Can anybody help me please?

I'm having a problem with (what I'm almost certain is) variable scope, maybe you can help...

So I've got a parent class that, among other things, gets the name of the filepath, explodes this filepath, checks for files with names that mirror that filepath (without the slashes of course), and includes them if they exist:



class ParentClass {

  public $RootFolder;
  public $PageSections;

  function __construct() {
global $root_folder;
$this->RootFolder = $root_folder;
//get all directory names...
$cleanse = $this->RootFolder;
$pagepath = str_replace($cleanse, '', dirname($_SERVER['PHP_SELF']));
$this->PageSections = explode('/', $pagepath);
  }

  function RequireIfExists ($sections, $root_folder, $file_name, $file_type) {
foreach ($sections as $i){
if (isset($o)) {
$i = $o."/".$i;
}
$file = $root_folder.$i."/".$file_name.$file_type;
if (file_exists($_SERVER['DOCUMENT_ROOT'].$file)) {
require ($_SERVER['DOCUMENT_ROOT'].$file);
echo "<!--".$file." included -->"; //temporary measure to ensure that the file is at least being called.
}
$o = $i;
}
  }
// so, excuting the function
// $this->RequireIfExists($this->PageSections, $this->RootFolder, "sectionvars", ".php");
// on some/sub/folder/index.php
// will look for somesectionvars.php, somesubsecitonvars.php & somesubfoldersectionvars.php
// and include whichever ones exist.
}

Then there is a child class that utilizes this function, and successfully requires the file. I know the file is coming through because the echo statement on the page shows up in the code, but I cannot get the variables to pass through?


<?php
class HtmlHead extends PageBlock{
public $PageKeywords;
public $PageDescription;
public $PageStyles;
public $PageScripts;
//------------------------
public $SectionKeywords;
public $SectionDescription;
//------------------------
public $SiteKeywords;
public $SiteDescription;
public $SiteStyles;
public $SiteScripts;
//------------------------
public $CatchAll;

  function __construct() {
parent::__construct();
//$this->RequireIfExists($this->PageSections, $this->RootFolder, "sectionvars", ".php");  //I have tried calling the function here with no luck
global $site_description, $site_keywords, $section_description, $section_keywords, $xyz; //also made sure to global the variables from the required page
$this->SiteDescription = $site_description;
$this->SiteKeywords = $site_keywords;
$this->SectionDescription = $section_description;
$this->SectionKeywords = $section_keywords;
$this->PageDescription = $page_description;
$this->PageKeywords = $page_keywords;
  }

function Constructor() {
$this->RequireIfExists($this->PageSections, $this->RootFolder, "sectionvars", ".php"); //Tried calling the function here as well, no luck
global $xyz; //global'd the var i want
echo "\n<meta http-equiv='Content-Type' content='text/html; charset=UTF-8' />";
echo "\n<!--Sec Keywords are".$xyz."-->"; //Tried to echo the var...
//------------------------Rest of code...
}
}


And here is the page that is being required...


<?php
echo "<!-- 1234 1234 1234 1234 -->"; //test measure to see if/where file is including
$section_keywords = "red, secondary";
$section_description = "this is the red section description";
$xyz = "1234"; //test Var to see if there was a naming conflict with the above two..
?>


What am I overlooking this time?   

Hello,   I recently posted here about an issue I was having with my database orientated products page.   I have now run into another problem where say if, /db.php was typed or /db.php?p=IDoNotExist was typed, it returns blank.   I have in my code the desired content to be displayed, but it just doesn't seem to want to make a show.   I was also wondering if it is possible to show different content for whatever the URL is, so for no parameter, the content about the products, and a non existent one, maybe "Product not found"?   Here is my code:

<?php
	$db=mysql_connect  ("localhost", "webwibco_charlie",  "Hello123") or die ('I cannot connect to the database  because: ' . mysql_error()); 
	$mydb=mysql_select_db("webwibco_products"); 

	include("header.php");

	$status = htmlspecialchars( @$_GET ['p'] );

	if ($status == "floorpuzzles") {

		     echo "<h1>Our Floor Puzzles</h1>";
		$sql="SELECT  ID, Name, Tags, Description, Category FROM products WHERE Category LIKE '%" . FloorPuzzles .  "%'"; 
		$result=mysql_query($sql); 
		while($row=mysql_fetch_array($result)){ 
			$Name  =$row['Name'];
			$ID  =$row['ID']; 
			$Description  =$row['Description'];  
			echo "<div class=\"box\">";
			echo "<h1>$Name</h1>";
			echo "<div class=\"floorbox\"><a href=\"?p=$ID\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"small\"></a></div>";
			echo "<h2>$Description</h2>";
			echo "</div>";
		}
?>
<?
}else{



	if ($status == $_GET["p"]) {
		$sql="SELECT  ID, Name, Tags, Description, Pieces, Size, Barcode, Category FROM products WHERE ID = '" . $_GET['p'] .  "'";   
		$result=mysql_query($sql);   
		while($row=mysql_fetch_array($result)){  
			$Name  =$row['Name'];
			$ID  =$row['ID']; 
			$Description  =$row['Description']; 
			$Pieces  =$row['Pieces']; 
			$Size  =$row['Size']; 
			$Barcode  =$row['Barcode'];   
			echo "<div class=\"1\">";
			echo "<h1>$Name</h1>";
			echo "<div class=\"bigbox\">";
			echo "<div class=\"floorbox\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"big\"></div>";
			echo "</div>";
			echo "</div>";
			echo "<div class=\"2\">";
			echo "<p>Puzzle Pieces: $Pieces</p>
				  <p>Puzzle Size: $Size</p>
				  <p>Barcode: $Barcode</p>";
			echo "</div>"; 
		} 
		}else{
?>
<?
			echo"<h1>Our Products</h1>
				<p>Our jigsaw puzzles are hand cut by skilled craftsmen and therefore each one is unique with self-correcting pieces. There is a strict quality control process at all stages by our highly experienced staff. The puzzles are durable and provide fun and excitement, enhancing learning and a child’s development.<p>
				<p>All of our jigsaws are made using materials from sustainable resources grown in managed forests. Where possible we support companies in the UK and source our components locally, most of our suppliers are in the East Midlands, many in Derbyshire and Nottinghamshire. We keep packaging to a minimum and take our environmental and ethical responsibilities very seriously.</p>
				<p>Reducing waste and recycling was a way of life for us before it became fashionable. We are constantly searching for new ideas and consult teachers when developing our jigsaws, which are often used within the national curriculum.</p> 
				<p>As well as making our own range, we manufacture for leading suppliers to the education market. Check for \"Made in Britain\" and it is probably made by us.</p>
				<p>We have a wide variety of products available for viewing, from classic floor puzzles to innovative inset trays. You can take a look at all our products on this page, simply use the navigation buttons to your left.</p>";

				
				}}

include("footer.php");
?>

The final echo is what I wish to be displayed on the URL without or with an invalid parameter.   Here is my site URL: http://www.webwib.co...saws/search.php (note that only the "Floor Puzzles" category has content within it).   Thank you in advance for assistance.

im creating a members website and i want to show how many people are logging, in my database i have a col named online every time somone logs in there online goes from 0 to 1 and when thay log out it goes back to 0.

i need to know how to show the total people that have 1 in there online part of the data base iv tryed this code and its not working

<?php
$result = mysql_query("SELECT online FROM `members` WHERE online='1'");

$row = mysql_fetch_row($result);

echo $row;
         ?>

this works in the sql console in phpmyadmin

Hi there, i am working on a mobile site can u pls help me with a script which shows guests online on my site with their country flag next to its phone model

Hi, i'm trying to create some detailed statistics about customer activity, i have entries in my mysql db when the customer has been active for the last time and want to create some statistics about that, basically a "online last 24 hours" but from specific countries.

Now i've tried this:

Code: [Select]

$time = date('Y-m-d H:i:s');
$time24 = date("Y-m-d H:i:s", time()-((60*60)*24));
$query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de' AND time BETWEEN '$time' AND '$time24' ";

to get the current date and select all customers that have been available from the current date minus 24 hours, what's my mistake here, as this doesn't seem to work!

Thanks

"vijdev and 0 Guests are viewing this topic."

can someone help me with a pointer on how to get started with something like the above message when viewing different pages in the site...it must be page/topic specific.
is this using sessions/cookies/databases?...?

i got my user login and register working with my sql but now if a non logged in user tries to access the shoutbox i want it to redirect them to the register page.

<?PHP

session_start();

if (!(isset($_SESSION['login']) && $_SESSION['login'] != '')) {
header ("Location: /login/main.php");
}

?>



im using that code but even if im logged in, it redirects me to the register page? 

my main site is on the root of the site. the login page and the logged in page is in "/login/

im trying to count and display the number on users on my site this is the coding im using cant see where im going wrong, its inserted into the data base correctly but wont delete after 60 seconds,

cheers matt



$session=session_id();
$time=time();
$time_check=$time-60;

$sql="SELECT * FROM onlineusers WHERE session='$session'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($count=="0"){
$sql1="INSERT INTO onlineusers(session, time, username)VALUES('$session', '$time', '$username')";
$result1=mysql_query($sql1);
}
else {
"$sql2=UPDATE onlineusers SET time='$time' WHERE session = '$session'";
$result2=mysql_query($sql2);
}

$sql3="SELECT * FROM onlineusers";
$result3=mysql_query($sql3);

$count_user_online=mysql_num_rows($result3);


$sql4="DELETE FROM onlineusers WHERE time<$time_check";
$result4=mysql_query($sql4);

Can somebody put a code that would show all the people that are online on the website?

ok well i have this is online script that starts at the login page where it sets  a session. well it echos that that person is online even if they are not, i will have all of the code only for the online script so it will be in peaces.
ok so here is the login page where the session is started
login.php

Code: [Select]
<?php
//ok so if they submit the page and its all right and they login , here is the session that is set for the person, again its just the piece of the script .
$_SESSION['logedin'] = $_POST['email'];
                    mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$id'");
?>
now here is where i call on the session and see if there online , the profile.php is set up to where it sees if it is your profile or not so $logoptions_id is the id that they are logged in as.
profile.php
Code: [Select]
<?php
// this is on top of the page , where the session is called and if they are logged in it updates the database where there id is. 
if( isset($_SESSION['logedin']) ) {
mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$logOptions_id'");// there is where $logOptions_id comes in.
}
// now this is further down the page(script) where we see if they are logged in or not.
$age= 60; //set a variable called age, assign an integer of 60 to it.
if( isset($_SESSION['logedin']) ) { 
$q = 'SELECT id=`$id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$id` <> \''.($_SESSION['logedin']).'\''; 
$isonlinecheck = mysql_query($q);
$row = mysql_fetch_assoc($isonlinecheck); 
if (($row['last_activity_stamp'] + $age)< time()){
$isonline =  "is <font color='green'>online!</font>";} 
else {
$isonline = "is<font color='red'> offline!</font>";
}

}
?>
i wana thank all who helps! your all greatly appreciated

I want to know how can i check wether the user is online on my website or not....

Hiya,

What's the best way of finding out and displaying the total number of users currently online, along a breakdown of how many of these users are members or guests?

I also wish to find out and display the total number of users who were online today, along with a breakdown of how many of these users are members or guests.

I've noticed that users of the same computer could of course use a different browser preference to another user of the same computer.
What should happen in this situation?

Any help is much appreciated.

Cheers!

ok i want to make it to where when you look at there profile it will tell you if they are online or not!
how would i do that ? here is my php script that has some modifications on the profile page but wont show if other users are online! here is the login script where i put the session and the profile page.
login.php
Code: [Select]

<?php



// Start Session to enable creating the session variables below when they log in

session_start();

// Force script errors and warnings to show on page in case php.ini file is set to not display them

error_reporting(E_ALL);

ini_set('display_errors', '1');

//-----------------------------------------------------------------------------------------------------------------------------------

// Initialize some vars

$errorMsg = '';

$email = '';

$pass = '';

$remember = '';

if (isset($_POST['email'])) {

    

    $email = $_POST['email'];

    $pass = $_POST['pass'];

    if (isset($_POST['remember'])) {

        $remember = $_POST['remember'];

    }

    $email = stripslashes($email);

    $pass = stripslashes($pass);

    $email = strip_tags($email);

    $pass = strip_tags($pass);

    

    // error handling conditional checks go here

    if ((!$email) || (!$pass)) { 



        $errorMsg = '<font color="red">Please fill in both fields</font>';



    } else { // Error handling is complete so process the info if no errors

        include 'scripts/connect_to_mysql.php'; // Connect to the database

        $email = mysql_real_escape_string($email); // After we connect, we secure the string before adding to query

        //$pass = mysql_real_escape_string($pass); // After we connect, we secure the string before adding to query

        $pass = md5($pass); // Add MD5 Hash to the password variable they supplied after filtering it

        // Make the SQL query

        $sql = mysql_query("SELECT * FROM myMembers WHERE email='$email' AND password='$pass' AND email_activated='1'"); 

        $login_check = mysql_num_rows($sql);

        // If login check number is greater than 0 (meaning they do exist and are activated)

        if($login_check > 0){ 

                while($row = mysql_fetch_array($sql)){

                    

                    // Pleae note: Adam removed all of the session_register() functions cuz they were deprecated and

                    // he made the scripts to where they operate universally the same on all modern PHP versions(PHP 4.0  thru 5.3+)

                    // Create session var for their raw id

                    $id = $row["id"];   

                    $_SESSION['id'] = $id;

                    // Create the idx session var

                    $_SESSION['idx'] = base64_encode("g4p3h9xfn8sq03hs2234$id");

                    // Create session var for their username

                    $username = $row["username"];

                    $_SESSION['username'] = $username;

                    //THIS IS WHERE I EDITED THE SESSION TO SAY IF THERE LOGGED IN OR NOT

                    $logedin = $row['id'];

                    $_SESSION['islogedin']=$logedin;

                    mysql_query("UPDATE myMembers SET last_log_date=now() WHERE id='$id' LIMIT 1");

                    // THIS WAS JUST A TEST BUT WONT UPDATE UNTILL THEY LOGOUT

                    mysql_query("UPDATE myMembers SET online='online' WHERE id='$id' LIMIT 1");

                    

                    

                } // close while

    

                // Remember Me Section

                if($remember == "yes"){

                    $encryptedID = base64_encode("g4enm2c0c4y3dn3727553$id");

                    setcookie("idCookie", $encryptedID, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days

                    setcookie("passCookie", $pass, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days

                } 

                // All good they are logged in, send them to homepage then exit script

                header("location: home.php?test=$id"); 

                exit();

    

        } else { // Run this code if login_check is equal to 0 meaning they do not exist

            $errorMsg = "<h3><font color='red'>Email/Password invalid<br /></font></h3><a href='forgot_pass.php'>Forgot password?</a><div align='right'> <br> Forget to activate you account?</div>";

        } 

        



    } // Close else after error checks



} //Close if (isset ($_POST['uname'])){



?><!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" />

<link rel="icon" href="favicon.ico" type="image/x-icon" />

<link rel="shortcut icon" href="favicon.ico" type="image/x-icon" />

<link href="style/main.css" rel="stylesheet" type="text/css" />

<script src="js/jquery-1.4.2.js" type="text/javascript"></script>

<title>Log In</title>

<title>Login Page</title>

   <style type="text/css">

        #stage {

            top: 0px;

            left: 0px;

            z-index: 100;

        }

        .stage {

            position: absolute;

            top: 0;

            left: 0;

            width: 100%;

            min-width: 900px;

            height: 1359px;

            overflow: hidden;

        }

        #bg {

            background: #aedfe5 url(images/sky1.png) 0 0 repeat-x;

        }

        #clouds {

            background: transparent url(images/cloud.png) 305px 10px repeat-x;

        }

        #sun

        {

            background:  url(images/land_sun.gif)0 0 no-repeat;

        }

        #hillbottom

        {

            background: url(images/hill2.png)0 1270px repeat-x;

        }

    </style>

    <link rel="stylesheet" type="text/css" href="css/loginstyle.css" /></head>

<body>

<!-- IE6 fixes are found in styles/ie6.css -->

    <!--[if lte IE 6]><link rel="stylesheet" type="text/css" href="css/ie6.css" /><![endif]-->

   

    <script src="js/jquery-1.3.2.min.js" type="text/javascript"></script>

    <script src="js/jquery-ui-1.7.2.spritely.custom.min.js" type="text/javascript"></script>

    <script src="js/jquery.spritely-0.5.js" type="text/javascript"></script>



    <script type="text/javascript">

    (function($) {

            $(document).ready(function() {

               

               

                   

                var direction = 'left';

               

               



                $('#clouds').pan({fps: 40, speed: 0.5, dir: direction, depth: 10});

               

               

            });

        })(jQuery);

   

    </script><div id="bg" class="stage"></div>

    <div id="container">

    <div id="sun" class="stage"></div>

    <div id="clouds" class="stage">

        <div id="stage" class="stage">

<body>

<div id="behindform">

  <form id="signinform" action="login.php" method="post" enctype="multipart/form-data" name="signinform">

    <fieldset>

   

      <legend>Log in</legend>

     

      <label for="login">Email</label>

      <input type="text" id="email" name="email" />

      <div class="clear"></div>

     

      <label for="password">Password</label>

      <input type="password" id="password" name="pass" />

      <div class="clear"></div>

     

      <label for="remember_me" style="padding: 0;">Remember me?</label>

      <input type="checkbox" id="remember" style="position: relative; top: 3px; margin: 0; " name="remember"/ value="yes" checked="checked">

      <div class="clear"></div>

     

      <br />

     

      <input type="submit" style="margin: -20px 0 0 287px;" class="button" name="commit" value="Sign In"/>

    </fieldset><?php print "$errorMsg"; ?>

  </form>

  </div>

  </div>

  </div><div id="hillbottom" class="stage"> </div>

 

 



 

   

   

</body>

</html>

profile.php This is only a part where i try. but when i putt it on , it wont echo the other peoples on , like it doesnt get the other sessions or somethig
Code: [Select]

//HERE IS WHERE I STARTED , BUT dONT KNOW WHAT TO DO !

if (isset($_SESSION['islogedin']) && $logOptions_id != $id) {

    $isonline = "<font color='green'>online</font>";

}

else{

    $isonline = "<font color='red'>offline</font>";

    }



// This is to Check if user is online or not! needs editing

//$isonline = mysql_query("SELECT online FROM myMembers WHERE id='$logOptions_id'AND online='online'");

//$isonlinecheck=mysql_query($isonline);

//if ($isonlinecheck ="online"){

//$online = "is <font color='green'>online!</font>";}

//else {

//    $online = "is<font color='red'> offline!</font>";

//}

// End to Check if user is online or not!

?>