can you tell me whats wrong with this?

[Glenskie]

can you tell me whats wrong with this?

 

<?php
$age= 60;
if( isset($_SESSION['logedin']) ) {
$q = mysql_query('SELECT id=`$id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> "'.($_SESSION['logedin']).'"');
$isonlinecheck = mysql_query($q);
if ($isonlinecheck ="last_activity_stamp + $age "< time()){
	$isonline =  "is <font color='green'>online!</font>";}
else {
$isonline = "is<font color='red'> offline!</font>";
}

}
?>

[requinix]

What's wrong with... your post? Sure. You posted a chunk of code with a very vague question (the same one in your post as you used for the thread title, even). No background information. No explanation of the code. No description of the problem.

[sunfighter]

And this:

 

$q = mysql_query('SELECT id=`$id`, .......

$isonlinecheck = mysql_query($q);

 

Makes no sense.

[Glenskie]

ok sorry , its check the database for the time and where the id is placed ,  it keeps showing online no matter what

[jcbones]

Not really sure how the logic behind this works, but I understand the problems with this script.

 

Why yes, I would be glad to break it down for you.

 

<?php
$age= 60; //set a variable called age, assign an integer of 60 to it.
if( isset($_SESSION['logedin']) ) { //if a session index called logedin is set, it could be empty though (most likely is, because the session has NOT been started).
$q = mysql_query('SELECT id=`$id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> "'.($_SESSION['logedin']).'"'); //run a mysql_query (I don't see a database connection here), then store the Resource in a variable call q.
$isonlinecheck = mysql_query($q); //attempt to assign a Resource of a Resource to another variable called isonlinecheck (it won't work).
if ($isonlinecheck ="last_activity_stamp + $age "< time()){ //run an if statement which reads: set a string of 'last_activity_stamp + 60' to a variable called isonlinecheck, then see if that string is less than the current timestamp.
	$isonline =  "is <font color='green'>online!</font>";} //set a string to a variable called isonline.
else {
$isonline = "is<font color='red'> offline!</font>";
}

}
?>

 

So, there is a few problems here.

 

1. You must start a session (if auto session start is not enabled on the server), at the top of the script.

2. You must connect to a database.

3. You are calling the query and storing the result resource, then calling another query on that.  I assume that was a mistake, (hey it happens).

4. In the if statement you use a single =, which is an assignment operator, comparison operators is == and ===.

5. Also in the if statement I think you wanted to add age to the last_activity_stamp, but you did it in string syntax which will NOT work.

 

So, I will take my suggestions and incorporate them into the script.  Crossing my fingers and hoping that I get it right on the first time around (I doubt it, I'm all jacked up on 5 hour energy ATM).

 

WHOOOOHOOOOOOO.

 

<?php
session_start();
/*Please add your database connection details below*/
include('database_connection_file.php');
/*database connection should have already been made*/

$age= 60; //set a variable called age, assign an integer of 60 to it.
if( isset($_SESSION['logedin']) ) { //if a session index called logedin is set, it could be empty though (most likely is, because the session has NOT been started).
$q = 'SELECT id, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> \''.($_SESSION['logedin']).'\''; 
$isonlinecheck = mysql_query($q); //Assign a Resource to a variable called isonlinecheck (it won't work).
$row = mysql_fetch_assoc($isonlinecheck); //get the row of data from the Resource.
if (($row['last_activity_stamp'] + $age)< time()){ //run an if statement
	$isonline =  "is <font color='green'>online!</font>";} //set a string to a variable called isonline.
else {
$isonline = "is<font color='red'> offline!</font>";
}

}
?>

 

I think that is it... if it doesn't work, post back and I'll probably still be up trying to work out all this 5 hour energy.

 

AHHHHHHHHHHHH....

[Glenskie]

hey i have to go to be but i will put the rest of the script up ... thank you for trying