Whos Online
People are wanting whos online scripts so here's a simple one.This idea came from Diablosblizz.
Run this in phpmyAdmin
Code
ALTER TABLE `members` ADD `online` INT( 1 ) NOT NULL DEFAULT '0';
In login.php find..
Code
$_SESSION['id'] = "$user[id]";
$_SESSION['password'] = "$user[password]";
And after put..
Code
$online = mysql_query("INSERT INTO `members` (`username`, `online`) VALUES('$username','1')");
This makes the user online as such. making the value 1 when they login and 0 when they logout.
So that makes the user online. but what if they wanted to logout?
In logout.php find..
Code
if(isset($_GET['logout'])){
And after put..
Code
$offline = mysql_query("INSERT INTO `members` (`username`, `online`) VALUES('$username','0')");
this shows them as offline
Want to view whos online. Nah didn't think so but here it is anyways.
Code
<?php
include ("config.php");
$sql = mysql_query("SELECT * from members where online == '1'");
while ($online = mysql_fetch_array($sql))
{
$username = $online[username];
echo ("- $username<br>");
//displays the online users
}
?>
this grabs all the users that have the value 1 which means they are online and displays it
be sure to look out for more tuts by be.
Posted on Saturday 15th November 2008 at 09:57 PM
this is right
<?php
session_start();
include"config.php";
$users = mysql_query("SELECT * FROM `members` WHERE `online`='1'");
while($online = mysql_fetch_array($users))
{
echo $online['username'] . " " . $online['ID'];
echo "<br />";
}
?>
A quick trip to W3SCHOOLS don't hurt anyone
<?php
session_start();
include"config.php";
$users = mysql_query("SELECT * FROM `members` WHERE `online`='1'");
while($online = mysql_fetch_array($users))
{
echo $online['username'] . " " . $online['ID'];
echo "<br />";
}
?>
A quick trip to W3SCHOOLS don't hurt anyone
Posted on Thursday 31st July 2008 at 03:46 PM
SQL query is wrong.
Posted on Thursday 31st July 2008 at 12:51 PM
I get this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/*****/public_html/HighRize/main/whosonline.php on line 126
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/*****/public_html/HighRize/main/whosonline.php on line 126
Posted on Tuesday 3rd June 2008 at 09:54 PM
Why is this code doing a insert instead of an update to set if the user is online or not?
Posted on Tuesday 3rd June 2008 at 08:16 PM
Alright well this is the 2nd tutorial where it has not worked simply because people just aren't testing it or are ripping it. I blow at coding and could see a problem from the start. Tutorial tells you to Insert into 'members' which simply only re-adds the name thats logged in with.
Im not sure how the bbcode on this comment system goes so i can only type the stuff out so here this goes in the login.php as stated in the tutorial:
$online = mysql_query("UPDATE `members` SET`online` = '1' WHERE `username` = '$username'");
This goes in the logout.php as shown above:
$offline = mysql_query("UPDATE `members` SET `online` = '0' WHERE `username` = '$logged[username]'");
And this is just a small add-on to where is shows whos online i linked it to the profile.
<?php
include("config.php");
$sql = mysql_query("SELECT * from members where online = '1'");
while ($online = mysql_fetch_array($sql))
{
$username = $online[username];
echo ("<a href='members.php?user=$username'>$username</a> ");
//displays the online users
}
?>
Im not sure how the bbcode on this comment system goes so i can only type the stuff out so here this goes in the login.php as stated in the tutorial:
$online = mysql_query("UPDATE `members` SET`online` = '1' WHERE `username` = '$username'");
This goes in the logout.php as shown above:
$offline = mysql_query("UPDATE `members` SET `online` = '0' WHERE `username` = '$logged[username]'");
And this is just a small add-on to where is shows whos online i linked it to the profile.
<?php
include("config.php");
$sql = mysql_query("SELECT * from members where online = '1'");
while ($online = mysql_fetch_array($sql))
{
$username = $online[username];
echo ("<a href='members.php?user=$username'>$username</a> ");
//displays the online users
}
?>
Posted on Tuesday 29th May 2007 at 01:29 AM
Well see that is the problem. There are bits of code to edit into each login and logout. So the only thing is you need to know is I followed this tutorial to a T.
Still only a blank page.
Still only a blank page.
Posted on Tuesday 29th May 2007 at 01:18 AM
post you code so we can take a look
Posted on Monday 28th May 2007 at 10:34 PM
Sorry bout the double post, but please help me in a PM.
Thanks
Thanks
Posted on Monday 28th May 2007 at 03:13 PM
All I see is a blank page when this is properly installed.
Help?
Help?
Posted on Monday 14th May 2007 at 11:11 PM
Nah, my idea doesn't even work after what you posted in forums....