[SOLVED] show users online

[MadTechie]

change

UPDATE `username` SET online='0' WHERE memberId='$row[id]' LIMIT 1

 

to

 

UPDATE `username` SET online='0' WHERE memberId='{$row['id']}' LIMIT 1

[alohatofu]

$time=time();
$timeout=time()+(60*60); #now+1hour
mysql_query("UPDATE `username` SET timestamp='$time' WHERE memberId='$id' LIMIT 1"); 
$query=mysql_query("SELECT * FROM `surv_member` WHERE `online`='1' AND `timestamp`>='$timeout'")
or die(mysql_error());
while($row=mysql_fetch_array($query)){
mysql_query("UPDATE `username` SET online='0' WHERE memberId='{$row['id']}' LIMIT 1"); 
$row['online'];
print "UPDATE `username` SET online='0' WHERE memberId='{$row['id']}' LIMIT 1";
}

//spit out the results
mysql_close();
print "{$row['username']}";

 

Still didnt' output anything.  On column online what type it should be? Binary?

[shaunrigby]

varchar(1) is easiest

[alohatofu]

bleh.. that shouldn't matter though right? I changed it and it didn't change anything.

 

I guess it's on the UPDATE

[shaunrigby]

Code revamp:

 

<?php

$time=time();

$timeout=time()+(60*60); #now+1hour
`
mysql_query("UPDATE `username` SET `timestamp` = '" . $time . "' WHERE `memberId` = '" . $id . "' LIMIT 1")
or die("Update Query 1: " . mysql_error()); 

$query = mysql_query("SELECT * FROM `surv_member` WHERE `online`= '1' AND `timestamp` >= '" . $timeout . '")
or die("SELECT Stament: " . mysql_error());

while($row=mysql_fetch_array($query)){
     mysql_query("UPDATE `username` SET `online` = '0' WHERE `memberId` = '" . $row['id'] . "' LIMIT 1")
or die("Update Query 2: " . mysql_error());

print $row['online'];
}

mysql_close();
?>

[shaunrigby]

What does it return?

[MadTechie]

try this

 

on every page ($id = userid i assumed that is set)

<?php
$time=time();
$timeout=time()+(60*60); #now+1hour
mysql_query("UPDATE `username` SET timestamp='$timeout' WHERE memberId='$id' LIMIT 1");

$query=mysql_query("SELECT * FROM `surv_member` WHERE `timestamp` >= '$time'") or die(mysql_error());
while($row=mysql_fetch_array($query))
{
$display[] = $row['username'];
}

//spit out the results
mysql_close();
foreach($display as $d)
{
print "$d<br />";
}
?>

 

 

logout

<?php
$time=time();
mysql_query("UPDATE `username` SET timestamp='$time' WHERE memberId='$id' LIMIT 1");

?>

 

No online field or login script required

[alohatofu]

Code revamp:

 

<?php

$time=time();

$timeout=time()+(60*60); #now+1hour
`
mysql_query("UPDATE `username` SET `timestamp` = '" . $time . "' WHERE `memberId` = '" . $id . "' LIMIT 1")
or die("Update Query 1: " . mysql_error()); 

$query = mysql_query("SELECT * FROM `surv_member` WHERE `online`= '1' AND `timestamp` >= '" . $timeout . '")
or die("SELECT Stament: " . mysql_error());

while($row=mysql_fetch_array($query)){
     mysql_query("UPDATE `username` SET `online` = '0' WHERE `memberId` = '" . $row['id'] . "' LIMIT 1")
or die("Update Query 2: " . mysql_error());

print $row['online'];
}

mysql_close();
?>

 

there are syntax errors somewhere on here. I couldn't get it to accept

[shaunrigby]

what line? print that line...

[MadTechie]

` on the 6th line

[alohatofu]

try this

 

on every page ($id = userid i assumed that is set)

<?php
$time=time();
$timeout=time()+(60*60); #now+1hour
mysql_query("UPDATE `username` SET timestamp='$timeout' WHERE memberId='$id' LIMIT 1");

$query=mysql_query("SELECT * FROM `surv_member` WHERE `timestamp` >= '$time'") or die(mysql_error());
while($row=mysql_fetch_array($query))
{
$display[] = $row['username'];
}

//spit out the results
mysql_close();
foreach($display as $d)
{
print "$d<br />";
}
?>

 

 

logout

<?php
$time=time();
mysql_query("UPDATE `username` SET timestamp='$time' WHERE memberId='$id' LIMIT 1");

?>

 

No online field or login script required

 

it doesn't like the foreach statement

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\sb\index.php on line 567

[shaunrigby]

what error message does my script return?

[MadTechie]

update

<?php
$time=time();
$timeout=time()+(60*60); #now+1hour
mysql_query("UPDATE `username` SET timestamp='$time' WHERE memberId='$id' LIMIT 1");

$query=mysql_query("SELECT * FROM `surv_member` WHERE `timestamp`>='$timeout'") or die(mysql_error());
$display = array();
while($row=mysql_fetch_array($query))
{
$display[] = $row['username'];
}

//spit out the results
mysql_close();
foreach($display as $d)
{
print "$d<br />";
}
?>

[shaunrigby]

Dude, we are swamping the kid with code, alohat, try my code, then post back th me the error message you are receiving, then do the same with madtechcie's code

[alohatofu]

what error message does my script return?

 

that syntax on the 6th line and there's something else on the script I'm still trying to figure out where is the syntax error =(

[alohatofu]

Shaun,

 

When i put your code I have

 

Parse error: syntax error, unexpected T_LNUMBER in C:\xampp\htdocs\sb\index.php on line 556

 

line 556 is

    mysql_query("UPDATE `username` SET `online` = '0' WHERE `memberId` = '" . $row['id'] . "' LIMIT 1")

 

[MadTechie]

missing ;

 

EDIT:

hate to ask but did my code work ?

[shaunrigby]

no, or die is on the next line

[alohatofu]

Madtech,

 

it didnt' give any error but it's not showing anything neither.

 

 

Shaun,

I'm still can't figure out where the syntax error is.. yes it's not missning the ; because "die" is on the next line

[shaunrigby]

try commenting out the second update statement and the or die beneath it, do you still get an error?

[MadTechie]

lol

would help if i check the logic a little closer

 

<?php
echo "ID = $id"; //Added for debugging
$time=time();
$timeout=time()+(60*60); #now+1hour
mysql_query("UPDATE `surv_member` SET timestamp='$time' WHERE memberId='$id' LIMIT 1");

$query=mysql_query("SELECT * FROM `surv_member` WHERE `timestamp`>='$timeout'") or die(mysql_error());
$display = array();
while($row=mysql_fetch_array($query))
{
$display[] = $row['username'];
}

//spit out the results
mysql_close();
foreach($display as $d)
{
print "$d<br />";
}
?>

 

ok done

[alohatofu]

commenting it still didn't take out the error

 

here's the image

[MadTechie]

line 552 is missing a )

[shaunrigby]

do you have

<?php
$server = "localhost"; // usually localhost
$db_user = "username"; 
$db_pass = "password"; 
$database = "survtask"; 
mysql_connect($server, $db_user, $db_pass);
?>

 

above it?

[shaunrigby]

it goes off the page