Text counter..

Posted: Thu Feb 20, 2003 11:34 am (21 years, 2 months ago)

I need a counter that will count, the users viewing my site currently, and the maximum amount of users on my site at one time, any suggestions?

I have PHP and MySQL access.
Thanks Smile

Posted: Thu Feb 20, 2003 5:48 pm (21 years, 2 months ago)

you could always find the relevant bit of code in a forum and somehow work that into your site...seems like a complicated thing to try to write from scratch Very Happy

Posted: Fri Feb 21, 2003 10:09 am (21 years, 2 months ago)

i don't have any experience with forums to do that, if anyone could find me such a code it would be very helpful Very Happy

Team Member Joined: 06 Jan 2002 Posts: 2564

Here are a few interesting links:

http://codewalkers.com/seecode/75.html
http://www.devarticles.com/art/1/122/3
http://php.inc.ru/main.php?view=tutorials&t=p_active

Posted: Sat Feb 22, 2003 11:10 am (21 years, 1 month ago)

wow, thanks!! Very Happy

Posted: Sat Feb 22, 2003 7:46 pm (21 years, 1 month ago)

http://www.gtchat.de/download_misc_en.html

Example @ http://www.gtchat.de/main_en.html

Posted: Wed Feb 26, 2003 8:21 pm (21 years, 1 month ago)

has anyone found a way to make a 'most online users' thing yet?

thanks for those guys!!

Team Member Joined: 06 Jan 2002 Posts: 2564

What is wrong with the links provided above? Do you mean that you require the "most users online" thing as well as "x users online"?

Posted: Wed Feb 26, 2003 8:32 pm (21 years, 1 month ago)

The links above are very helpful and i have used them and found a script i needed Smile

But i could also do with a most online users too, i understand i'm nagging you Sad

Team Member Joined: 06 Jan 2002 Posts: 2564

Which one did you use? We can't help you unless we know which one you chose Smile

Posted: Sun Mar 02, 2003 5:32 pm (21 years, 1 month ago)

I used this code :

Code:

<?php
$server = "localhost";
$db_user = "...";
$db_pass = "...";
$database = "...";

$db = mysql_connect($server, $db_user,$db_pass);
mysql_select_db($database,$db);

$timeoutseconds = 900; //5 minutes

$timestamp = time();
$timeout = $timestamp-$timeoutseconds;

$insert = mysql_query("INSERT INTO useronline VALUES ('$timestamp','$REMOTE_ADDR','$PHP_SELF')",$db);
if(!($insert)) {
print "Useronline Insert Failed > ";
}
$delete = mysql_query("DELETE FROM useronline WHERE timestamp<$timeout",$db);
if(!($delete)) {
print "Useronline Delete Failed > ";
}
$result = mysql_query("SELECT DISTINCT ip FROM useronline WHERE file='$PHP_SELF'",$db);
if(!($result)) {
print "Useronline Select Error > ";
}
$user = mysql_num_rows($result);

mysql_close();
if($user == 1) {
print("Users online: $user\n");
} else {
print("Users online: $user\n");
}
?>

Team Member Joined: 06 Jan 2002 Posts: 2564

OK, you should now create a table like this:

Code:

CREATE TABLE config (
config_name varchar(255) NOT NULL,
config_value varchar(255) NOT NULL,
PRIMARY KEY (config_name)
);

Then add some code to this effect:

Code:

$most_online = mysql_query("SELECT config_value FROM config WHERE config_name = 'most_online'",$db);

while ($row = mysql_fetch_array($most_online, MYSQL_ASSOC)) {
$row['config_value']=$most_online2;
}

if ($user>$most_online2) {
$update_online = mysql_query("UPDATE config SET config_value = $user WHERE config_name = 'most_online'",$db);
}

There's probably something wrong with this code, but the idea's there Very Happy

You'll have to add a row into the config table with:
config_name='most_online' and config_value=0.

Posted: Sun Mar 02, 2003 8:12 pm (21 years, 1 month ago)

should i put that table in a new database or just add to the one i have?

Team Member Joined: 06 Jan 2002 Posts: 2564

Add it to the one you have. Smile

Posted: Mon Mar 03, 2003 5:39 pm (21 years, 1 month ago)

Thanks!