PHP - Php If..showing A Different Status

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

Hi there, i have code that enables a user to login to my site. I have a player table that consists of: PlayerName, Password and Status. The playername and password is what logs the user in however i want to make it so that  when a user logs in it automatically sets that users Status to 1. Can some one point me in the right direction please??

Thanks

Code: [Select]
<?php if (!isset($_SESSION)) {
  session_start();
}

$loginFormAction = $_SERVER['PHP_SELF'];
if (isset($_GET['accesscheck'])) {
  $_SESSION['PrevUrl'] = $_GET['accesscheck'];
}

if (isset($_POST['textfield'])) {
  $loginUsername=$_POST['textfield'];
  $password=$_POST['textfield2'];
  $MM_fldUserAuthorization = "";
  $MM_redirectLoginSuccess = "index.php";
  $MM_redirectLoginFailed = "fail.php";
  $MM_redirecttoReferrer = false;
  mysql_select_db($database_swb, $swb);
  
  $LoginRS__query=sprintf("SELECT PlayerName, Password FROM player WHERE PlayerName=%s AND Password=%s",
    GetSQLValueString($loginUsername, "text"), GetSQLValueString($password, "text")); 
   
  $LoginRS = mysql_query($LoginRS__query, $swb) or die(mysql_error());
  $loginFoundUser = mysql_num_rows($LoginRS);
  if ($loginFoundUser) {
     $loginStrGroup = "";
    
    //declare two session variables and assign them
    $_SESSION['MM_Username'] = $loginUsername;
    $_SESSION['MM_UserGroup'] = $loginStrGroup;       

    if (isset($_SESSION['PrevUrl']) && false) {
      $MM_redirectLoginSuccess = $_SESSION['PrevUrl'];
    }
    header("Location: " . $MM_redirectLoginSuccess );
  }
  else {
    header("Location: ". $MM_redirectLoginFailed );
  }
} ?>
Thank you

Howdy
I'm trying to figure out to get a file to call a page on another domain(diffrent server) and check if the activation code matches the one on the activation server.

So on the user's server it has a cron job or on load will call my server and run the code in their update.php file against my activation.php file and see if it matches.

If($usr == $actv){
     //DO THIS
}else{
     //run disable script
}

Hello,

I am trying to figure out how to grab the latest aim status update from "http://lifestream.aol.com/stream/iragedxt"

But I can only see the Current Status if I am logged into my aim account on the website so is there a way I can log into my aim account on lifestream.aol.com and then I believe its to use curl to grab the text from the other website?

if so could someone please help me with this?

Thanks,
John

hello
when i try to get the checkbox status from my class i get only those who checked
i send the &_POST to my class.

how should i get also the values of checkboxes which not selected?

I got this code off of this forum
Code: [Select]
<?php
$ip="{$_GET['ip']}";
$port="{$_GET['port']}";
header('Content-type: image/png');
if(!$sock=@fsockopen($ip,"$port", $num, $error, 5)) {
readfile('image/offline.png');
} else {
readfile('image/online.png');
}
?>I'm able to check the status by going to Code: [Select]
http://domain.com/statuscheck.php?ip=google.com&port=80 and it works fine. But I would like to show the status on my index.html page without having to click on a link. Is there anyway Please help.
thank you in Advance.

Hello

I recently got a "to many connection" error on my site, and want to know if anyone here knows a few codes
that will show how many connection currently are in use (maybe even what files that creates them).

I found I can use "Threads_connected" to show current open connections, but no info on how to write the code or where to pu the
file.

I hope you can show my what to do

I am trying to make a Server Status checker that uses a image to tell if it is online or not. I also am trying to make it so you can the URL as a image code(EX: http://Http://domain.com/statuscheck.php?ip=google.com&port=80). What I'm trying to say it, that if you do like <img src="http://Http://domain.com/statuscheck.php?ip=google.com&port=80"> it would go to my site, use the php script, and then have the image Online or Offline, were the <img> code was posted.

Here's my code Im trying to make it from.
<?php
$ip="{$_GET['ip']}";
$port="{$_GET['port']}";
if(!$sock=@fsockopen($ip,"$port", $num, $error, 5)) {
Header('Content-type: image/png');
echo"<img src='images/offline.png'>";
} else {
Header('Content-type: image/png');
echo"<img src='images/online.png'>";
}
?>

Hi all, I have this script I used on a site hosted on my machine that checked if different servers and applications were running on my machine.
Here's the code.

Code: [Select]
<html>
        <body>
        <style>
        fd{
color: #000000
font-size:100%
}
</style>

<?php
function check_port($port) {
    $conn = @fsockopen("212.233.xxx.xxx", $port, $errno, $errstr, 0.2);
    if ($conn) {
        fclose($conn);
        return true;
    }
}

function server_report() {
    $report = array();
    $svcs = array('21'=>'FTP',
                  '22'=>'SSH',
                  '25'=>'SMTP',
                  '80'=>'HTTP',
                  '110'=>'POP3',
                  '143'=>'IMAP',
                  '3306'=>'MySQL',
                                  '8080'=>'HFS',
                                  '8081'=>'Webcam',
                                  '3784'=>'Ventrilo'); 
    foreach ($svcs as $port=>$service) {
        $report[$service] = check_port($port);
    }
    return $report;
}

$report = server_report();
?>

<center>
<table>

    <tr>
        <td><font face="Verdana" size="1" color="#000000">Apache:</td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['HTTP'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>
       
        <tr>
        <td><font face="Verdana" size="1" color="#000000">MySQL:</td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['MySQL'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>
       
    <tr>
        <td><font face="Verdana" size="1" color="#000000">HFS:</td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['HFS'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>
       
         <tr>
        <td><font face="Verdana" size="1" color="#000000">FTP:</font></td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['FTP'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>
       
        <tr>
        <td><font face="Verdana" size="1" color="#000000">Webcam:</font></td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['Webcam'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>
       
        <tr>
        <td><font face="Verdana" size="1" color="#000000">Ventrilo:</font></td>
        <td><font face="Verdana" size="1" color="#000000">
                <?php echo $report['Ventrilo'] ? "<img src='main/images/on.png' width='16' height='15' border='0'>" : "<img src='main/images/off.png' width='16' height='15' border='0'>"; ?></font></td>
    </tr>


</table>
</center>

</body>
</html>
I've added the HTML too. As you can see it checks if the ports are open. It worked perfectly and I didn't have any issues with it at all.
Now, for a couple of days, I've been using an awardspace paid hosting, bought a domain name.
I use wordpress now, and I pasted this code directly in a PHP/HTML code box (I use a plugin that enables use of PHP) and it shows the HTML formatting how it should be, but unfortunately doesn't work as intended.
It shows everything to be down, while apache, mysql and my ftp server are still running on my computer (the IP it checks, 212.233.xxx.xxx).
When I set the IP to be 127.0.0.1, the IP of the awardspace server as far as my script is concerned, it shows only that the IP is working.

So, any suggestions? Any idea would be greatly appreciated 

So i have a online status available and when the user logs in i insert a new row in a table so the user is online and when the user logs out i delete it so the users goes offline.. but when the user leaves the computer and has the website open he will get logged out after some time of inactivity and the row saying that the user is online will remain there and wont be deleted.. how can i remove the row by the time the session is removed due to inactivity?

I dont have any code done till now because im still trying to find out how to do it..

thanks.

Hey Guys/Girls, Thanks for offering to help!

I'm currently setting up a small social network for school and I just basically want to know whether the way I'm dealing with in-active users is appropriate and not going to SLOW down my code A LOT,
My method is :

3 Functions:

Code: [Select]
function update_active_user(){
global $connection;
global $id;
$time = time();
$result3 = mysql_query("UPDATE users5 SET last_update = '$time' WHERE id = '$id'", $connection);
}
function update_inactive_users(){
global $connection;
$time_to_expire = time() - 300; // 300 seconds off the current time
$result2 = mysql_query("UPDATE users5 SET online='0' WHERE last_update < $time_to_expire AND online='1'", $connection);
}

function update_active_users(){
global $connection;
$time_to_expire = time() - 300; // 300 seconds off the current time
$result2 = mysql_query("UPDATE users5 SET online='1' WHERE last_update > $time_to_expire AND online='0'", $connection);
}

First function updates the user's field called last_update to the current time
(This will only occer when the script loads and the user has done something on my website, it will update their last_update field to current time)

Second function sets ALL users that haven't loaded any pages in the last 300 seconds to offline or field online to 0, which obviously means the user is offline

Third function sets ALL users that have loaded pages in the last 300 seconds to online if they're offline

The reason I'm worried about this is not because it doesn't work, it's working fine at the moment with 2 users, I'm worried when their might be A LOT of users and there are people who are active.

My question is : will it make my scripts really slow or be bad for my database in ANY way if I do use these functions for ALL users because each time it searches the table, it is searching ALL users to see if they're active or inactive.

I have an AJAX request but my problem is that is always returns a status code of 0.  Below I have my JavaScript function.  They send a request to a page called inc_submit.php which just echo's "hi" (as I'm just try to test it). 

Code: [Select]
function submitForm(website_name, website_caption, website_contact, website_email, website_link) {
var url4 = "includes/inc_submit.php?website_name=" + escape(website_name) + "&website_caption=" + escape(website_caption) + "&website_contact=" + escape(website_contact) + "&website_email=" + escape(website_email) + "&website_link=" + escape(website_link);
alert (url4);
request4.open("GET", url4, true);
request4.onreadystatechange = confirmSubmit;
request4.send(null);
}

function confirmSubmit() {
if (request4.readyState == 4) {
if (request4.status == 200) {
var response4 = request4.responseText;
document.getElementById("submit_status").innerHTML = response4;
} else {
alert("Error: status code is " + request4.status);
}
}
}

If i check more than one checkbox. I dont get any value from database.   HTML

<form method="post">
    <input type="checkbox" name="receiver[]" value="cheater">CHEATER
    <input type="checkbox" name="receiver[]" value="un-verified">UN-VERIFIED
    <input type="checkbox" name="receiver[]" value="inactive">INACTIVE
    <input type="checkbox" name="receiver[]" value="active">ACTIVE
    <input type="submit" name="submit" value="ADD">
</form>

PHP

$errors = array();
$success = NULL;
$error = NULL;
$var['receiver'] = isset($_POST['receiver']) ? $_POST['receiver'] : NULL;
.................

if(!empty($_POST['submit'])){
    // FORM VALIDATION //
}

if(!empty($_POST['submit']) and empty($errors)){
	
    $status = array_map('strval', $var['receiver']) + array(0);
    $statusSql = implode(',', $status);
	
    $query = 'SELECT email FROM users WHERE status IN("'.$statusSql.'")';
    $select = $db->prepare($query);
    $select->execute();
	
    $arrayData = array();
    while($row = $select->fetch(PDO::FETCH_ASSOC)){
        $arrayData[] = $row['email'];
    }
	
    $errors[] = implode(',', $arrayData);
	
}

require_once 'includes/antiCsrf/index.php';
$csrf = new antiCsrf();

$smarty->assign('success', $success);
$smarty->assign('error', $error);
$smarty->assign('errors', $errors);
$smarty->assign('csrfKey', $csrf->csrfKey());
$smarty->assign('csrfToken', $csrf->csrfToken());
$smarty->assign('var', $var);
.................

Hello, I made a server statuc checking script that checks the status of a game server. It works fine on my computer, but when I upload it to my web host it stops working. I'm assuming my web host is blocking it in some way. My host is Yahoo Small Businesses. Does anyone know if they block this kind of thing, or if there is a way to unblock it? Perhaps editing the code?

Here is my code, if that helps.
Code: [Select]
<?php
function GetServerStatus($site, $port)
{
$status = array("OFFLINE", "ONLINE");
$fp = @fsockopen($site, $port, $errno, $errstr, 2);
if(!$fp)
{
return $status[0];
}
else 
{
return $status[1];
}
}
$Redemption = GetServerStatus("174.142.118.233", 11661);
$RedemptionPossible = GetServerStatus("174.142.118.234", 51665);

$Tranquility = GetServerStatus("174.142.118.234", 11601);
$TranquilityPossible = GetServerStatus("174.142.118.234", 51665);

$Nemesis = GetServerStatus("174.142.118.241", 11601);
$NemesisPossible = GetServerStatus("174.142.118.241", 64721);

$OverallServer = GetServerStatus("174.142.118.237", 11093);
$CodServer = GetServerStatus("8.9.6.122", 4230);
echo($CodServer . "<br />");
echo("Overall Server: " . $OverallServer . "<br />");
echo("Redemption: " . $Redemption . "<br />");
echo("Tranquility: " . $Tranquility . "<br />");
echo("Nemesis: " . $Nemesis . "<br />");
?>

i need to pull the records from where status not = rejected does anyone know the correct syntax?
Code: [Select]
$data=mysql_query("SELECT * FROM design WHERE id='$id' AND status!='rejected' ") or die (mysql_error());