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PHP - User, Online, Offline Status
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. Similar TutorialsI use firebug to detect how javascript work. It same code. the java script :
<script type="text/javascript">
$(function() {
$("#cmbNegara").change(function(){
$("img#imgLoad").show();
var kodejeniskayu = $(this).val();
$.ajax({
type: "POST",
dataType: "html",
url: "getProvinsi.php",
data: "kodejeniskayu="+kodejeniskayu,
success: function(msg){
if(msg == ''){
$("select#cmbProvinsi").html('<option value="">--Pilih Ukuran--</option>');
$("select#cmbKota").html('<option value="">--Pilih Ukuran--</option>');
}else{
$("select#cmbProvinsi").html(msg);
}
$("img#imgLoad").hide();
getAjaxAlamat();
}
});
});
$("#cmbProvinsi").change(getAjaxAlamat);
function getAjaxAlamat(){
$("img#imgLoadMerk").show();
var kodeukuran = $("#cmbProvinsi").val();
$.ajax({
type: "POST",
dataType: "html",
url: "getKota.php",
data: "kodeukuran="+kodeukuran,
success: function(msg){
if(msg == ''){
$("select#cmbKota").html('<option value="">--Stok Kosong--</option>');
}else{
$("select#cmbKota").html(msg);
}
$("img#imgLoadMerk").hide();
}
});
};
});
</script>
getprovinsi.php
<?php require_once('Connections/karyo.php'); ?>
<?php
ini_set('display_errors',0);
//ambil parameter
$kodejeniskayu = $_POST['kodejeniskayu'];
if($kodejeniskayu == ''){
exit;
}else{
$sql ="SELECT kodeukuran,ukuran FROM tukuran WHERE kodejeniskayu='$kodejeniskayu'";
$getNamaProvinsi = mysql_query($sql,$karyo) or die ('Query Gagal');
while($data = mysql_fetch_array($getNamaProvinsi)){
echo '<option value="'.$data['kodeukuran'].'">'.$data['ukuran'].'</option>';
}
exit;
}
?>
getkota.php
<?php require_once('Connections/karyo.php'); ?>
<?php
ini_set('display_errors',0);
//ambil parameter
$kodeukuran = $_POST['kodeukuran'];
if($kodeukuran == ''){
exit;
}else{
$sql ="SELECT idstok,harga FROM tstok WHERE kodeukuran = '$kodeukuran'";
$getNamaKota = mysql_query($sql,$karyo) or die ('Query Gagal');
while($data = mysql_fetch_array($getNamaKota)){
echo '<option value="'.$data['idstok'].'">'.$data['harga'].'</option>';
}
exit;
}
?>
the problem at online is, the combobox at getprovinsi not showing up.
I want to know how can i check wether the user is online on my website or not.... 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. What is the best way to keep track of a User's (real-time) online status? Right now I have a Log In script and I set the status in the User's Session, and then "pass" that status from page to page with session_start();. However, the problems with this approach include... 1.) A User is idle for hours or days and thus not really online 2.) A User leaves my site and thus is not really online (Or does closing the window change that?) I want a fairly accurate and real-time reading of whether or not a User is actively using my site so I can set their "Online Status" icon as green (online) or yellow (in active) or grey (offline). Any idea how to accomplish that so I have monitoring like you'd see on an Instant Messenger App? Thanks, Debbie I've got a site that requires the user to follow a well-defined workflow, corresponding to a sequence of pages. You'll have the right idea if you think of a store where the user selects items, then chooses a delivery method, then pays (although the site's actual purpose is somewhat different than that). I just tracked down a puzzling logical failu a user followed the workflow up to a certain point, then opened a new browser window, pointed it to the site, and started a second workflow. The site stores information about the user's activities in session variables, and the two windows shared a single session, so the site got tremendously confused about what the user was doing. Given the site's design, visiting the site in two windows at once is a legitimate thing for a user to do. I'm darned if I see how I can accommodate it, though. Other designers must have dealt with this problem many times. What ways of solving it have been found effective? In this case, the simplest and most effective solution would be to create a new session for each window in which the user visits the site. Is there any way to do that? Code what i made so far. Your comments at what should i do differently.
My configs.php
<?php
$userQuery = 'SELECT * FROM users WHERE id = :id';
$user = $db->prepare($userQuery);
$user->bindParam(':id', $_SESSION['userId'], PDO::PARAM_INT);
$user->execute();
$userInfo = $user->fetch(PDO::FETCH_ASSOC);
?>
functions.php
<?php
function loginCheck(){
global $db;
if(isset($_SESSION['userId'], $_SESSION['loginString'])){
$query = 'SELECT username FROM users WHERE id = :id';
$user = $db->prepare($query);
$user->bindParam(':id', $_SESSION['userId'], PDO::PARAM_INT);
$user->execute();
$row = $user->fetch(PDO::FETCH_ASSOC);
if($user->rowCount() == 1){
if(hash('sha512', $row['username'].$_SERVER['HTTP_USER_AGENT']) == $_SESSION['loginString']){
return true;
}else{
return false;
}
}else{
return false;
}
}else{
return false;
}
}
function checkUserRole(){//can be user, admin and moderator
global $userInfo;
if($userInfo['userRole'] == 'admin' or $userInfo['userRole'] == 'moderator'){
return true;
}else{
return false;
}
}
?>
shoutbox.php
Can this be done with one query?
global $db, $userInfo; $sbQuery = 'SELECT * FROM shoutbox ORDER BY dateCreated DESC LIMIT 30'; $sb = $db->query($sbQuery); $usersQuery = 'SELECT * FROM users WHERE shoutBoxBan = "yes"'; $users= $db->query($usersQuery); $usersRow = $users->fetch(PDO::FETCH_ASSOC);
$hiddenAction = '';
while($sbRow = $sb->fetch(PDO::FETCH_ASSOC)){
if(loginCheck() and checkUserRole()){
$hiddenAction = " <a href=\"javascript:;\" onClick=\"deleteMessage('".$sbRow['id']."')\" class=\"shoutBoxDelete\" title=\"Delete\">x</a>";
if($usersRow['username'] == $sbRow['username']){
$hiddenAction .= " <a href=\"javascript:;\" onClick=\"unBan('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Unban\">u</a>";
}else{
if($userInfo['username'] != $sbRow['username']){//admin and moderator cant ban themselves.
$hiddenAction .= " <a href=\"javascript:;\" onClick=\"banUser('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Ban\">o</a>";
$hiddenAction .= " <a href=\"javascript:;\" onClick=\"tempBanUser('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Temp Ban\">ΓΈ</a>";
}
}
}
....................................
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 Can somebody put a code that would show all the people that are online on the website? 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 { // $online = "is<font color='red'> offline!</font>"; //} // End to Check if user is online or not! ?> i got my user login and register working with my sql but now if a non logged in user tries to access the shoutbox i want it to redirect them to the register page. <?PHP session_start(); if (!(isset($_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); it displays is online no matter what! please help me! here is the login.php where the session is started . 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'); //----------------------------------------------------------------------------------------------------------------------------------- include 'scripts/connect_to_mysql.php'; // Connect to the database // 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 $email = mysql_real_escape_string($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 $_SESSION['logedin'] = $_POST['email']; mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$id'"); mysql_query("UPDATE myMembers SET last_log_date=now() WHERE id='$id' LIMIT 1"); // THIS WAS JUST A TEST BUT WONT UPDATE UNTILL THEY LOGOUT } // close while // Remember Me Section if($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", $pass, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days } // All good they are logged in, send them to homepage then exit script header("location: /socialtscripts/home.php?test=$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>and then this is the profile.php where i see if there online , i could only put sertian areas , script is too big profile.php Code: [Select] <?php //on top of the page where it checks the session and updates the time // This updates the database correctly if( isset($_SESSION['logedin']) ) { mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$logOptions_id'"); // this is where it selects the users id but it wont work , it says online for every user! $age= 60; if( isset($_SESSION['logedin']) ) { $q = mysql_query('SELECT id=`$logOptions_id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> "'.($_SESSION['logedin']).'"'); $isonlinecheck = mysql_query($q); if ($isonlinecheck ="last_activity_stamp" + $age < time()){ $isonline = "is <font color='green'>online!</font>";} else { $online = "is<font color='red'> offline!</font>"; } } ?> PLEASE HELP ME Hey, guys. I'm a programmer and i've designed an application to assist in an online game. When users visit my website for whatever reason i want to display the amount of users currently using my application. As you may have guessed i'm not to crash hot at php or any web related languages at that . What i think will work, is when my application starts up, i can make it visit a webpage in the background and just idle there while they are using the program, when they close the program obviously it is no longer viewing that webpage. So if there some sort of php counter i can make for this? I was reading some sources to php counters and they count the amount of "sessions" and display that, i was just curious if my program navigates to a webpage upon opening and it doesn't do anything will this 'time it out'? Anyway, any help is appreciated. Thanks. I am using MAMP on my MacBook and don't understand why MAMP and my code in NetBeans will not run if I am offline? If I am trying to execute PHP files locally on my laptop, why should MAMP or Apache of NetBEans care if I do not have an Internet connection?! I wanted to show someone something at work where there is no Internet access, but that won't work as it currently stands. TomTees I have a self-built blog website, done with PHP and MySQL.
How to write the posts offline? How is it done? Which tools are used?
Any suggestions?
Hi, I need to know the delete status of an email. I can able to make the Read status, but whether there is any tricks that we can make in order to know the status of our email (deleted from our users) Any help !! Ty JO Problem: I'm trying to make it so the array responds to the integer of status in the database. So, to check if it's one, I try this: $get["status"]["1"] = "Message here..."; But, on my result, I'm just getting WWTT - And the value in the database for the ticket is 0. I'm not sure if I'm using the array correctly. My full code: <?php session_start(); include("../includes/mysql.php"); include("../includes/config.php"); ?> <!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=iso-8859-1" /> <link href="../style.css" rel="stylesheet" type="text/css" /> <title><?php echo $title; ?></title> </head> <body> <div id="container"> <div id="content"> <div id="left"> <div class="menu"> <?php include("../includes/navigation.php"); ?> <div class="menufooter"></div> </div> <?php include("../includes/menu.php"); ?> </div> <div id="middle"> <?php $existing = $_POST['existing']; $query = mysql_query("SELECT COUNT(id),id,status,date,question,title FROM tickets WHERE id='$existing'"); $get = mysql_fetch_assoc($query); if(!$existing) { echo ' <div class="post"> <div class="postheader"><h1>Error</h1></div> <div class="postcontent"> <p>You have not enetered in a ticket ID. Please go back and do so.</p> </div> <div class="postfooter"></div> </div> '; } elseif($get['COUNT(id)'] < 1) { echo ' <div class="post"> <div class="postheader"><h1>Error</h1></div> <div class="postcontent"> <p>The ticket ID you are trying to use doesnt exist. Please go back or submit another ticket.</p> </div> <div class="postfooter"></div> </div> '; } else { $get["status"]["0"] = "Waiting for support..."; $get["status"]["1"] = "Waiting for user..."; $get["status"]["2"] = "Ticket Closed..."; $get["status"]["3"] = "Ticket Opened..."; echo ' <div class="post"> <div class="postheader"><h1>View Ticket Status - ID '. $get["id"] .'</h1></div> <div class="postcontent"> <p>Title: '. $get["title"] .' - Posted on: '. $get["date"] .'</p> <p>Ticket Status: '. $get["status"] .'</p> <p>Question: '. nl2br($get["question"]) .'</p> </div> <div class="postfooter"></div> </div> '; } ?> </div> </div> </div> </body> </html> Hello, I have a webpage that runs some php code to control electrical devices. One part of it controls outside lights. Another part controls an electric door striker. I keep expanding the functionality of it and need to change one pin of the port without affecting the others. So far I use a different port for each function but I am out of ports so I need to use the different pins of the parallel port. I am using a linux ubuntu server. From my php webpage I issue this command: <?php exec("/home/setSerialSignal /dev/ttyS0 0 0"); ?> which controls the lights using the serial port. I can control 2 pins but I can't check the status of them so if I make a change to one pin it cancels the other. For another one I use the parallel port with this command: <?php exec("/home/parashell 0x378 1"); ?> which turns on pin #2 of the parallel port. Parashell has an executable called "pin". If you type the command "pin 0x378" it will return a value for the status of the parallel port. "1" for example which tells me that pin #2 is on. Is there a way to pass this value to a php variable? I am looking to be able to read the status of the port and make changes to just the pin I want. Lets say pin #2 is on and I want to turn pin #3 on. I would have to issue <?php exec("/home/parashell 0x378 2"); ?> to turn pin #3 on, but that would turn off pin #2. If I could pass the "1" from the "pin" command I could add it to the "2" and give it the command with a variable as "3". Thanks in advance for any help you can give. I hope I explained it ok and didn't confuse everyone. 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 I am wanting to echo one of these 4 statements depending on the 'status' value. Currently it is showing the status value on the 2nd line but not the statement below.. Code: [Select] <li> <strong>Status: </strong><?php more_fields('status') ?> <?php if (more_fields('status')=="Red") echo "Your account is currently undergoing judgement"; elseif (more_fields('status')=="Green") echo "Your account is currently Live"; elseif (more_fields('status')=="Yellow") echo "Your account is currently undergoing site visits"; elseif (more_fields('status')=="Blue") echo "Your account is currently undergoing insolvency/liquidation";?> </li> any help appreciated. Thanks, Jake |
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