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February 3, 2011
Answered

Friends feed application

  • February 3, 2011
  • 1 reply
  • 562 views

Hi Guys

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

{

}

This topic has been closed for replies.
Correct answer

bregent wrote:

Seems like you've asked this before.

The duplicate post can be seen here:

http://forums.adobe.com/thread/771787

1 reply

Participating Frequently
February 3, 2011

Seems like you've asked this before. Do you really have more than one database you are querying? Are they both on the same host? You should not need to execute multiple queries inside loops. Just do a simple SQL join in a single select statement. Is there a reason you are not doing it this way?

Correct answer
February 3, 2011

bregent wrote:

Seems like you've asked this before.

The duplicate post can be seen here:

http://forums.adobe.com/thread/771787

February 3, 2011

can you help?