PHP - Php Mysql Friend's Online Script Help

I currently making a relatively simple turn-based strategy game (kind of like a rock, paper, scissors game) in flash and I have no idea as to what I should use as a socket (or like a socket) to relay players' interactions from one client to another.  I have researched various APIs like nonoba and SmartFoxServer, etc but they either force their own interface on you or charge an arm and a leg.  I've read that java would achieve what I want, but I just don't know enough about java, sockets or networking to make sense of any of it, let alone trying to utilize it.

So I am thinking about using php and a mysql database since I am familiar with them, but I'm concerned about performance even though the game would not be very demanding as it is basically just passing values from client to client via php/mysql.

Little info on the game:
3-6 players per room. I have no idea as to how many rooms would be necessary, of course that will be determined by its popularity. The timing is what will concern me, as it must be accurate -1 min intervals or less if all of the players make their selections sooner, but they must all be in sync. I only see about 6-10, small (<20 characters/per) variables needing to be passed per player - so if it were 10 variables, in a room with six players, player 1 would send 10 variables and receive 50 back
Another option, that I admittedly haven't researched yet, is an xml socket, but I have no idea how to accomplish this and am concern with performance with that as well - probably just because I'm ignorant about it. 

So what do you guys think?  Think php/mysql could handle this?  Or do I need to learn another language/API?  Any input would be greatly appreciated.

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.

Is it possible to simply fetch from the server(mySql) and then dump it online through a PHP page?
Is there some link that you can share?

Hello All,

I'm making a networking script for my cms to allow users to befriend each other, I've got adding each other as friends down, that is no longer an issue.
My issue is displaying friends on a profile page.

Say Matt added Frank, and frank accepted the request, well on Matts page it'll show frank as a friend, but on franks page it won't show Matt as a friend.

My MySQL table has just 4 Rows

id | user_id | friend_id | approved

-id is a primary unique key
-user_id is who sent the reuqest
-friend_id is the id of the user the request was sent to
-approved is the status, 0 for no accepted, 1 for accepted.

On the user profile page, this is the sql query used to generate the friends list

$friends = $db->get_table("SELECT * FROM zxt_friends WHERE user_id = '{$u['id']}' AND approved = '1'");
foreach ($friends AS $friend)
{

$friend['friend_name'] =$zext->user_cache[$friend['friend_id']]['username'];

$avatar = $zext->user_cache[$friend['friend_id']]['avatar'];
        $friend_html = $friend['friend_name'].$avatar;
}


Before I tried using a different sql query, it looked like this:
$friends = $db->get_table("SELECT * FROM zxt_friends WHERE user_id = '{$u['id']}' OR friend_id = '{$u['id']}' AND approved = '1'");

But the approved status was ignored, and it was all kinds of messed up, on my page it would show that Matt is indeed friends with frank, and a member that had not approved the reuqest yet, and on Frank's page it showed that he was a friend with himself when that wasn't even the case.

I've looked at array_push but I can't seem to figure out the query or what I need to add the the loop, any help would be much appreciated.

Thanks, Matt.

This topic has done the Monster Mash to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=343718.0

Hi everyone,

I built a php site which uses $_GET['id'] to create a page for every picture in a gallery.

I added a Google Friend Connect(http://www.google.com/friendconnect) widget which is suppost to allow ratings and comments on every page but every page is treated as index.php

page examples

http://daroom.info/index.php?id=2053345520104164875S600x600Q85.jpg

http://daroom.info/index.php?id=21.jpg

Does anyone have any solutions to this problem?

I have made a mutual friend system like facebook and it shows all the mutual friends but sometimes it shows it twice because your 2 friends might be friends with one person so it echos the same person twice so i wanna only show the person once and the most repeated should show on the top and the less repeated person should show at bottom!

Right i want to have friends displayed on peoples profiles but i can only ave there name and not the avater displayed here is the code ive got which doesnt work

        <? 
$picture = mysql_query("SELECT * FROM users WHERE username = '$dip->person'");
$pc = mysql_fetch_object($picture);
$query_friends=mysql_query("SELECT * FROM friends WHERE username='$viewuser' AND type='Friend'");

$rows=mysql_num_rows($query_friends);
if ($rows == "0"){ echo "<center>No friends</center>"; }
$friend = 0;
while($dip=mysql_fetch_object($query_friends)){

echo "&nbsp;&nbsp;<img src='$pc->image' width='50' height='50' border='1'><br><a href='profile.php?viewuser=$dip->person'>$dip->person</a>,"; 
$friend++; 
echo ($friend % 3 == 0)? "<br>" : ""; }
?>


The friends are from the friends database and the avater is from the users database how can i link them so the name and the avater show ive tried this but only the names are displayed and the avaters dont show

Right ive got a user profile that i want a add friend button but i coded a little something what i fort wud work but no luck
<?php
session_start();
include "includes/db_connect.php";
include "includes/functions.php";
include"includes/smile.php";
logincheck();
$username=$_SESSION['username'];
$viewuser=$_GET['viewuser'];

$fetch=mysql_fetch_object(mysql_query("SELECT * FROM users WHERE username='$viewuser'"));
if (!$fetch){
echo "No such user";


$totalf = mysql_num_rows(mysql_query("SELECT * FROM friends WHERE username = '$viewuser' AND active='1'"));

$invite_text="<div>$username Has Sent You A Friend Request<br>
<input name=Yes_Accept type=submit id=yes value=Accept Invite class=abutton>&nbsp;<input name=No_accept type=submit value=Decline Invite class=abutton></div><input type=hidden name=invite_id value=$bar2>";
if (($_GET['fri'])){
$exicst=mysql_query("SELECT * FROM users WHERE username='$viewuser'");
$nums=mysql_num_rows($exicst);
$adding=mysql_fetch_object($exicst);
$already=mysql_num_rows(mysql_query("SELECT * FROM friends WHERE type='Friend' AND person='$viewuser' AND username='$username'"));
if ($already != "0"){
echo "<center><font color=orange><b><br>This user is already your friend.<br><br></font>";
}elseif ($already == "0"){


mysql_query("INSERT INTO `friends` ( `id` , `username` , `person` , `type` , `active`) 
VALUES (
'', '$username', '$viewuser', 'Friend' , '0'
)");
mysql_query("INSERT INTO `friends` ( `id` , `username` , `person` , `type` , `active`) 
VALUES (
'', '$viewuser', '$username', 'Friend' , '0'
)");
mysql_query("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `subject` , `date` , `read`) 
VALUES (
'', '$viewuser', '$username', '$invite_text' , 'Friend Request' , '$date' , '0'
)");
$bar2=mysql_insert_id();
echo "<center><font color=orange><br>Your Friend Invitation Was Sent To $viewuser<br><br></font>";
exit();
}
}}
?>
<a href=?fri=Yes>Add Friend +</a>

It just adds a blank person and comes back with No Such User and Your Friend Invitation Was Sent To

I think ive put some things in the wrong place to be honest but as im not a pro i easily miss things

Need some help here!

I want to build a friendgroup - I have a full working friendsystem.

My ide is this. I got a list of friends, make a group named Test. Put selected friends in there, make another group and put another friends in there.

Some idees how to do that with prepared statements??

I search for tutorials, but couldn't find any. Any suggestions?

The functions I need is this:

1. Make group
2. Add friend to group
3. Show group with selected friends
4. Rename groups
5. Move friend from one group to another
6. Delete group with all friends inside
7. Delete a empty group
8. Option to search for friends inside a group

Tutorial, free source code - everything would help to solve this mystery !

So far is this the code I made, and didn't work at all

public function create_friendgroup($profileownerid, $friends = NULL, $name)
{

   $sql = "SELECT fg_friends_id FROM friend_groups WHERE fg_member_id = '$profileownerid' LIMIT 1";

      if($stmt = $this->conn->prepare($sql))
      {
         $stmt->execute();
         $stmt->bind_result($friend);
         $stmt->fetch();
         $stmt->close();

      }

   if($friend != null)
      {
         $ua = explode(",", $friend);
         $ua = array_unique($ua);

         foreach ($ua as $u)
         {
         if($u !=NULL && $u !=$profileownerid)
         {
         $oldusers .= "{$u},";
      }   }
   }

$newusers = $profileownerid;
      if ($oldusers != null)
         {
         $newusers .= "," . $oldusers;
         }

$sql = "INSERT INTO friend_groups(fg_member_id,fg_friends_id, fg_name, fg_created_date) VALUES (?,?,?,?)";

$date = date("d-m-Y H:i");
      
      if($stmt = $this->conn->prepare($sql))
      {
         $stmt->bind_param('iiss',$profileownerid,$friends = trim($newusers, ","), $name,$date);
         $stmt->execute();
         $stmt->close();
      }
}







K.F

Hi,

Can anyone help me with a script, I have tried a if statement but I cannot get it to work and its driving me mad.

Basically I have a string my website uses to get if the user is logged in what their username is and I want it so when they click a certain link it checks to see if a table all ready exists called their username, if it does it displays a message, if it doesnt it creates the table and if the username is "anonymous" is displays a message.

So in short:

if $username is the same as a table display "Table all ready exists"
if $username="anonymous" display "You must be logged int"
if $username not the same as a table then create table.

Many thanks in advance

Jay

TRUNCATE TABLE CubeCart_cats_idx;
INSERT INTO CubeCart_cats_idx (productid, cat_id) SELECT productid, cat_id FROM CubeCart_inventory;

I an running the following in MyPHPAdmin, but need this to be automated in a PHP script, how would I do it ?
all help is appreciated, still new to PHP & MySQL 

many thanks

D

Hello,
I'm new to this website so forgive me if I'm posting in the wrong section.
I just need to know if there is a way that I can retrieve php script that is stored in a database?

for example, in my database in table:
<? echo"Hello World"; ?>

and in my php file, I call that using:
<?
mysql_connect("asdas","asdas","asd");
   
mysql_select_db("asdasda");

$result = mysql_query("select * from somedatabse");

while($r=mysql_fetch_array($result))
{   

   $table=$r["table"];
   
}

 echo"$table";
?>

i know its confusing, i just hope i made sense and that someone understands what im trying to do.

any help will be appreciatedddd
thank you : )