PHP - Online Friends Script

I've been looking for a simple script that would connect to a game server to see if it is still online. I've found many, but not one of them work. They will permanently send back the text "Online", or "Offline". This is the simplest code I have found:


<?php
$ip = "66.79.190.40"; 
$port = "27960"; 
if (! $sock = @fsockopen($ip, $port, $num, $error, 5)) 
echo '<B><FONT COLOR=red>Offline</b></FONT>'; 
else{ 
echo '<B><FONT COLOR=lime>Online</b></FONT>'; 
fclose($sock); 
} 
?>

I'm also looking into getting it to pull the map name and player-list, but I should be able to figure that out on my own once I get this working. If it helps, this script is supposed to connect to the original Quake games(One, two, and three).

I almost posted this in the freelancing forum, because I am looking for it to be created, but I first needed more info on what a coder would need to know about the project.

I have a script that is designed for a single person, and I want a coder to design it to become an online service with user registration, etc.

I'm weary of posting publicly the details about what the script does, but is something like that needed in order for a php programmer to take on a job? Could something like that be left out, until actually giving the coder the job?

Do I have to give at least a vague idea of what the script does?

I don't understand enough about what a programmer would need to know, but at the same time, I don't want to reveal my idea either.

Can somebody help me on how much I am going to need to reveal?

Thanks.

Table friends
users     
Fred
Tom
Julie
Henry
Bill
Wally
Joe
Joe
friendwith
Joe
Joe
Joe
Joe
Joe
Joe
Julie
Wally
Fred, Tom, Julie, Henry, Bill, and Wally have friended Joe.
Code: [Select]
<?php
$username = 'Joe';
$query = "SELECT * FROM friends WHERE friendwith=' $username'";
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
    echo $row['users']."<br/>";
    echo $row['friendwith']."<br/>";
}
?>   

Joe is friends with Julie and Wally.
Code: [Select]
<?php
$username = 'Joe';
$query = "SELECT * FROM friends WHERE users='$username'"; 
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
    echo $row['users']."<br/>";
    echo $row['friendwith']."<br/>";
}
?>
Here's where I need your expert advice!
I want to know who friended Joe except for those that Joe has friended.
In other words, I want to show Fred, Tom, Henry, and Bill.
Julie and Wally should not be included in the query because Joe has friended them as well.

Hi guys, im making a basic social networking site for a college project and I need to know how to make a system that will let logged in users add each other as friends but I carnt really find anything on google about how to start. Can anyone here help me or know a tutorial I can follow that will let me create something like this?

Thanks

Say i was going to store a "friends list" for a user. Would it be easier to store the names in a text string in the db instead of having to make a new field for each friend? In other words:

001 | Keith | Joe, Jane, Mike

or

001 | Keith | Joe | Jane | Mike

It seems like a pain to make a column for each but since i'm only 6 months into php i haven't really tapped into it's ability to deal with a string of data. My end game would to at some point take the user list and make a selectable table:

Friends:
Joe
Jane
Mike

Preferably with an array pseudo coded as:

Count how many users
for each user
echo username <a href username's page>



Bottom line: Store as Columns or String?

p.s. i've reviewed php.net etc. so it's not a matter of not knowing how but more which is preferable

Hi guys,

I'm developing a website which allows people to connect and follow each other's activity (like Twitter, for example).

To simplify everything, let's say I only have 2 tables:

1. Followers

id | follower_id | id_to_follow
------------------------------------

2. Activity

id | member_id | message | time
-----------------------------------------

Let's say John is following Jane and Bob. I want to create a "news" page and display the last 20 messages from Bob and Jane, chronologically.  For small numbers, I'd do something like this:

Select everything from the Activity table, check for every entry if the member is a friend of John's (in the Followers table) and, if so, display the message, ORDER BY `id` DESC.

But, this is very inefficient, I guess, for larger numbers (I can't even think about how many queries would take to do this on a site like Twitter...).

Any ideas of how to do the same thing more efficiently?

Thank you.

I have just a general question about a friends system here with regards to the database design.......

lets say you have a website where users can rent books.  The database design would be


MEMBER                                    BOOK                                                              LINk
memberid                                  isbn                                                                 memberid
name                                         name                                                              isbn
                                               
Or something to that extent

Now with a friends system like facebook has that "LINK" table would be absolutley huge and I cant see it being good design practise.  I mean how do you suppose facebook does this. Do they have ......

FRIENDS
myID
friendID


and just millions upon millions of repeated data?

 The only way I can see would be in the member table have a field called friends and then the ID's of each friend like so


Friends
001, 002, 003, 004, 005, 006, 007,

Now is my logic correct here or am I thinking of this totally the wrong way?  Would the other way be a better option or would it take the system too long to get to the 1000,000 th record in the database?

I believe the first option would be correct but I cant see how the server would handle the request if say you had 1000,000 users.  It just seems to me that you would create too many records. 

In my website people can sign up to mulitple events but if every user had 100 events the link table would just be massive!!!!!!!!!!!!!!!!!!!!!!

Im just after peoples input really on the situation.

Thank you

I'm trying to make it so i can show the friends i added, or users that added a friend


<table border=2 width="250" height="125"><tr>
<?php
$userfinal = get_username($_SESSION['user_id']);
$Members = mysql_query("SELECT * FROM friends WHERE username='$userfinal' AND friendname");
$numRowsMembers = mysql_num_rows($Members);

for($count = 1; $count <= $numRowsMembers; $count++)
{
    $name = mysql_fetch_array($Members);


?>
<td width="150" height="125">
<a href="view_profile.php?username=<? echo $name['friendname']?>"><img src="<? echo $name['main_P']?>" width="100" height="100"/>
<? echo $name['friendname']?></a>

<?
if (isset($name['date']) && (time() - $name['date'] > 300)) {

echo 'offline =[';
} else {

echo "<font color=green>[Online Now!]</font>";
}
$name['date'] = time(); // update last activity time stamp
?>

</td>




<? } ?>
</tr></table>



database


id  friendname  username
12  kristybellexo  zhshero
13  demo  zhshero
14  zhshero  zhshero

I have a database with all users.. Within that database all users have an id. How would I make it so users can "friend" other users? Would I need to make a new table for that?

Im sorry for the dumb question.. I havent dealt with php or mysql in about two years because I was busy getting engaged and all.. But now that I have time again I think im gonna take up one of my old projects. Granted I probably could have answered this question myself back then but now im finding some trouble taking up programming again. :\

Hi all

 

This part is integral to my site and im panicking as i cant seem to find an answer...

 

When someone clicks a link on my page with the below code it only does it as services... how can i create a paypal link to charge as friends and family PLEASE PLEASE PLEASE HELP

<form id="payid" name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post"> 
<input type="hidden" name="cmd" value="_xclick"> 
<input type="hidden" name="business" value="<?php echo $row1['EMAIL']; ?>"> 
<input type="hidden" name="currency_code" value="GBP">
<input type="hidden" name="item_name" value="BALL"> 
<input type="hidden" name="amount" value="<?php echo preg_replace( '/[^0-9,"."]/', '',  $row['PRICE'] ); ?>">
<!--<input type="hidden" name="return" value="http://www.theeasypc.co.uk/lottery/heandal.php">-->
<input type="hidden" name="return" value="http://www.theeasypc.co.uk/lottery/heandal.php?success=1">
    <input type="hidden" name="cancel_return" value="http://www.theeasypc.co.uk/lottery/heandal.php?error=1">

    <!-- Where to send the PayPal IPN to. -->
    <input type="hidden" name="notify_url" value="http://www.theeasypc.co.uk/lottery/heandal.php" />
<input type="image" src="http://www.paypal.com/en_US/i/btn/btn_buynow_LG.gif" border="0" name="submit" alt="Make payments with PayPal - it's fast, free and secure!">
</form>

I can see the  its set to business in one flag but ive tried to change thie to f&f, personal, friends, family, nothing works also tried "_donations"

 

 

Here is the error code I am receiving;   Parse error: syntax error, unexpected $end (line 41)   The error message has consistently identified the statement in the else clause.  Thanks for any help.

<?php session_start(); ?>

   <!DOCTYPE html PUBLIC "-//W3C//DD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

       <head>
       
	 <meta http-equiv="Content-Type" content="text/html; charset=utf-8"  />
         <title>Blah</title>
         <link rel="stylesheet" type="text/css" href="defaultcss.css" />

       </head>

	  <body>
	
	      <div id ="wrapper">
	
	      <h2>Blah Forum Demo</h2>
	
	      <p>Creating basic login functionality</p>
	
	<?php
	
	if (!isset($_SESSION['uid'])) 
	{
		echo "<form action = 'login_parse.php' method ='post'>
		Username: <input type ='text' name='username' /> &nbsp;
		Password: <input type ='password' name='password' /> &nbsp;
		Submit: <input type ='submit' name='submit' value='Log In' />
		";
	}
	else
	{
		echo "<p>You are logged in as ".$_SESSION['username']." &bull; <a href='logout_parse.php'>Logout</a>;
	}
	?>
	
	</div>
	
	</body>

</html>		

Hi Guys

 http://www.phpfreaks.com/forums/Smileys/nrg_alpha/cool.gif
Cool

I have social networking application. I am trying to query a database to get the ids of all the users friends.  With the ids of all the users friends I am then trying to  query a second database with  the status feed  that contains all users status data.

 

I would like to echo the users friends status data only from the second database

 

code is below hope you can help.

 

Thank you.

 

<?Php
session_start();
?>
<?Php
//connect.
include("connect.php");
//Time ago coverting code.
include_once("classes/develop_php_library.php"); // Include the class library

 

$id=$_SESSION['id'];

 

/*find-out users friends*/
$findperson=mysql_query("SELECT * FROM friends WHERE sessid='$id'");

 

$timeAgoObject = new convertToAgo; // Create an object for the time conversion functions

 

$findfriend=mysql_num_rows($findperson);

 

//Count if the person has any friends. If they have friends get the ids of all their friends
if($findfriend>0)
{
while($rati=mysql_fetch_assoc($findperson))
{
    $fried=$rati['friendid'];

 

    //query the status table to give the users friends status.

 

    $mediafeeds=mysql_query("SELECT * FROM status WHERE userid='$fried' LIMIT 0,8");

 

    $media_num=mysql_num_rows($mediafeeds);
}

 

//count to see if their any status updates from users friends.

 

if($media_num>0)
{

 

//display all the users friends status data.
$datamedia="<TABLE BORDER='0' CELLPADDING=8 bgcolor='#FFFFFF' align='center' width='350px' height='30px'>";

 

while($mini=mysql_fetch_assoc($mediafeeds))
{
$user_id=$mini['userid'];
$viewer_nme=$mini['username'];
$viewer_picture=$mini['viewerpics'];
$media_pic=$mini['contentpic'];
$desc_ption=$mini['description'];
$date_time=$mini['date'];

 

$convertedTime = ($timeAgoObject -> convert_datetime($datetime)); // Convert Date Time
$datetime = ($timeAgoObject -> makeAgo($convertedTime)); // Then convert to ago time

..

//This is just a table with the data of all the users friends data.


$datamedia.="<tr><td valign='top' cellpadding='5' width='10%' bgcolor='#FFFFFF' align='center' >$viewer_nme<br/><a href='friendsprofile.php?uid=$uidd&&viewer=$id'><img src='".$mediapic."' width='80' height='80' align=left></td>
      <td valign='top' align='left' cellpadding='5' width='60%' bgcolor='#D3D3D3' cellpadding='0'>$introduction<br/>$titlenamed$titled<br/>$descd$desc_ ption<br/><br/>$datetime</td></tr>";

 


    }
$datamedia.="</TABLE>";
echo $datamedia;
}
else
{
echo "<font color='#333333' size='2' face='sans-serif' align=left><div align='center'> Your friends have not current activities.</div></font>";
}
}

else

{

 

 

}

I am trying to make a filter to show posts from mutual friends between you and the persons profile you're on, I have somewhat of an idea on how to write the mysql query (it needs to be just one query, not two(which would be the easy way of doing it)).

the "friends" table has:
"friend_1" and "friend_2"

and the "posts" table:
"to_id" and "from_id"

The PHP variables would be:
"$session" and "$id"
id would be the users Id of the profile you're on.

Thanks, any help would be appreciated.

I'm working on  a Social Network from scratch. I'm stuck on how to store friend relationships between users?
On one topic here, I found an idea.
Have a table with 3 columns; id,user_1,user_2
each relationship would have a new row in the database.
Would this be the best way to do this? I can imagine the table would get quite big.
If each user has 100 friends and there's 10000 users that's 1000000 rows. etc

im creating a members website and i want to show how many people are logging, in my database i have a col named online every time somone logs in there online goes from 0 to 1 and when thay log out it goes back to 0.

i need to know how to show the total people that have 1 in there online part of the data base iv tryed this code and its not working

<?php
$result = mysql_query("SELECT online FROM `members` WHERE online='1'");

$row = mysql_fetch_row($result);

echo $row;
         ?>

this works in the sql console in phpmyadmin