Users friends status data for social application.

[T-Vision]

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

 

{

 

 

 

 

 

}

[BlueSkyIS]

excellent! any questions?

[T-Vision]

Well the question is I am finding it hard to query the second database with just the friendid variable from the first table. This is because the variable contain all the users friends ids as a single id. Making it hard to query the second database because friends id has become one number with all the users friends id.

 

 

Thanks