PHP - Next 16 Friends, And Previous 16 Friends Mysql Query Help

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 have a website where users can add their friends. What I am trying to achieve is to show every user that who are their friends that are currently online. If any one knows a link to any script or tutorial that explains this, Please help.

Thanks,

Faisal

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

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'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

I was just wondering if it's possible to run a query on data that has been returned from a previous query?  For example, if I do

Code: [Select]
$sql = 'My query';
$rs = mysql_query($sql, $mysql_conn);

Is it then possible to run a second query on this data such as

Code: [Select]
$sql = 'My query';
$secondrs = mysql_query($sql, $rs, $mysql_conn);

Thanks for any help

I'm trying to figure out how to filter the results of a database query.
For example somebody uses a php/javascript form to search for ford cars. When they see 100 results, they then narrow the results by model and/or color.

Am I right in thinking that the way to do this is by creating a temporary table of the results? What is the best approach to what I'm trying to do?

Hi all,
I have a fairly simple problem, but i cannot get to it. I have a page, that displays one (1) record at the time, and I want to have forward and back buttons (links) at the top and bottom of the page. Can anyone help me with that???
Here is a part of my code.

Code: [Select]
<form method="get">
<table border="0" width="90%" cellpadding="2" cellspacing="2" align="center">
<tr><td align="center">
  Select first letter or kind of the story:
  <input type="hidden" name="link" value="stories.php">
<input type="hidden" name="Pripadnost" value="3">
<select name="letter" onchange="this.form.submit()">
<?PHP
$letter =  trim($_REQUEST["letter"]);
?>

<option value="Sve" <?PHP echo $letter == "All" ? " selected " : "" ; ?>>All</option>
<?PHP
include_once ("/includes/databaseMySQL.php");
mysql_set_charset('utf8');

$SQL = "Select distinct ucase(left(title,1)) as sl FROM tblStoryes order by sl";

$rs_data = mysql_query($SQL) or die(mysql_error());
while ($row_data = mysql_fetch_assoc($rs_data)) {
      ?>
      <option value="<?PHP echo $row_data["sl"]; ?>"<?PHP echo $letter == $row_data["sl"] ? " selected " : "" ;
  echo ">".$row_data["sl"] ;?></option>
      <?PHP
 // ******************  Listed all the first letters of the titles into a drop box. *********************** //
} ?>
</select> &nbsp; &nbsp; &nbsp;
Bosnian <input type="checkbox" name="Bosnian " <?PHP echo isset($_REQUEST["Bosnian "]) ? "checked=checked" : "" ;?> onclick="this.form.submit()" />  &nbsp;
Turkish <input type="checkbox" name="Turkish "  <?PHP echo isset($_REQUEST["Turkish "]) ? "checked=checked" :  "" ;?> onclick="this.form.submit()" />  &nbsp;
Arabian <input type="checkbox" name="Arabian "  <?PHP echo isset($_REQUEST["Arabian "]) ? "checked=checked" : ""  ;?> onclick="this.form.submit()" />

</td></tr>
</table>

<table border="0" width="90%" cellpadding="2" cellspacing="2" align="center">
<?PHP
$id = -1;
$rs_data = mysql_query("Select max(id) as maxid from  tblStories WHERE aktivno = True") or die(mysql_error());
$row_data = mysql_fetch_assoc($rs_data);
$maxid = (int) $row_data["maxid"];

 // ******************  Found the largest ID from the table. *********************** //

$SQL = "SELECT * FROM tblStories WHERE aktivno = True ";


if ( strlen($letter) < 3 && strlen($letter) > 0 ) {
$SQL = $SQL . " and ucase(left(`title`, 1)) = '" . $Letter . "'";
}

$lang = 0;
$lang = $lang + isset($_REQUEST["Bosnian"]) * 4;
$lang = $lang + isset($_REQUEST["Turkish"]  ) * 2;
$lang = $lang + isset($_REQUEST["Arabian"] );
switch ($lang) {
    case 6: // bos i tur
        $SQL = $SQL . " and ( Bosanski = 1 or Turski = 1 ) ";
        break;
    case 5: // bos i ar
        $SQL = $SQL . " and ( Bosanski = 1 or Arapski = 1) ";
        break;
case 4: // bos
        $SQL = $SQL . " and Bosanski = 1 ";
        break;
    case 3: // tur i ar
        $SQL = $SQL . " and ( Turski = 1 or Arapski =1) ";
        break;
    case 2: // tur
        $SQL = $SQL . " and Turski = 1 ";
        break;
    case 1: // ar
        $SQL = $SQL . " and Arapski =1 ";
        break;
}


if (trim($_REQUEST["DBid"]) <> ""  or  $_REQUEST["DBid"] > 0)
$SQL = $SQL . " and  id > " . trim($_REQUEST["DBid"]) ;
$SQL = $SQL . " LIMIT 1"; //order by naslov

$rs_data = mysql_query($SQL) or die(mysql_error());
echo "<!--" . $SQL . " : Jezik je:" . $lang . "-->";
while ($row_data = mysql_fetch_assoc($rs_data)) {
$id = (int) $row_data["ID"];
?>
  <tr>
<td align="center" colspan=3>
       <h3><?PHP echo $row_data["Title"];?> </h3>
    </td>
    </tr>   
    <tr>
      <td width=20%>&nbsp;
      </td>
      <td align="justify" width=60%>
        <?PHP  echo $row_data["tekst"]; // tekst = text, but not keyword ?>
      </td>
      <td width=20%>&nbsp;
      </td>
    </tr>
<?PHP
}
?>
</table>
<p>

<?PHP
// ****************************** HEEEEELLLLPPPPPP ******************************* //
?>

<table border="0" width="90%" cellpadding="1" align="center">
  <tr>
    <td align="center"><a href="?DBid=1<?PHP
foreach($_REQUEST as $arr => $elem){
if ( $arr != "DBid") echo "&" . $arr . "=" . $elem;
} ?>"> &lt;&lt;</a></td>
    <td align="center"><a href="?DBid=<?PHP
echo $_REQUEST["id"];
foreach($_REQUEST as $arr => $elem){
if ( $arr != "DBid") echo "&" . $arr . "=" . $elem;
}
?>" >&lt;</a></td>
<?PHP if (  $maxid <=  $id ) { ?>
    <td align="center">&gt;</td>
    <td align="center">&gt;&gt;</td>
<?PHP }
  else
  {?>
    <td align="center"><a href="?DBid=<?PHP
echo $id ;
foreach($_REQUEST as $arr => $elem){
if ($arr != "DBid") echo "&" . $arr . "=" . $elem;
}
?>">&gt;</a></td>
    <?PHP if ( ($id + 9) <  $maxid ) { ?>
    <td align="center"><a href="?DBid=<?PHP
echo $id+10 ;
foreach($_REQUEST as $arr => $elem){
if ($arr != "DBid") echo "&" . $arr . "=" . $elem;
}
?>">&gt;&gt;</a></td>
    <?PHP
    }
    else
      {?>
        <td align="center">&gt;&gt;</td>
<?PHP }?>   
<?PHP }?>
  </tr>
</table>
</form>

I've got a page that displays all my blog entries on a single page. When you click on a specific entry, it pulls the post ID and loads the page specifically. From there, I wanted to have the option to click either onto the next or previous post by finding the next or previous ID in the database.. however, I know that sometimes posts are deleted, etc. so the ID's won't exactly be in order.

What direction should I take with this?

I want to loop through a MySQL table, and perform a calculation based on the current row the loop is on and the previous loop.

Say my table has 2 columns - Film_id and FilmRelease, how would I loop through and echo out a calculation of the current FilmRelease and the previous row's column value?

Thanks guys

I have got to this stage, but for some reason it's not printing out anything

Code: [Select]
<?php
 mysql_connect("localhost", "****", "*****") or die(mysql_error()); 
 mysql_select_db("*****") or die(mysql_error()); 

 $sql = mysql_query("SELECT FilmRelease FROM Films_Info") 
 or die(mysql_error()); 

 $last_value = null;

while ($row = mysql_fetch_assoc($sql)) {

  if (!is_null($last_value)) {

    print date_diff($row['FilmRelease'], $last_value) . "<BR>";
  }

  $last_value = $row['FilmRelease'];
}

Here is my code:
// Start MySQL Query for Records
$query = "SELECT codes_update_no_join_1b" .
"SET orig_code_1 = new_code_1,
       orig_code_2 = new_code_2" .
"WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)";
$results = mysql_query($query) or die(mysql_error());
// End MySQL Query for Records


This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this???

Code: [Select]
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
</div>
</div>
 
<div style="clear: both;"> </div>
</div>
 
<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>
 
</body>
</html>