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I am building my uncle an online music catalogue and am trying to use pagination http://www.phpeasycode.com/pagination/ but for some reason it is only showing the cat_id data from the databse and even then its only the last entry.

 

I need it to show all cat_id, song, artist and year from the database and not just the same ones over and over again.

 

Here is a link to the page so you can see whats happening: http://www.nealeweb.com/steven2/

 

And here is the code, Can anybody tell me where i have gone wrong please.

$rpp = 24; // results per page
$adjacents = 4;

$page = intval($_GET["page"]);
if($page<=0) $page = 1;

$reload = $_SERVER['PHP_SELF'];

// connect to your DB:
$dbHost = "";
$dbUser = "";
$dbPass = "";
$dbname = "";

$db = mysql_connect($dbHost,$dbUser,$dbPass);
mysql_select_db($dbname,$db);

// select appropriate results from DB:
$requete = "SELECT cat_id, song, artist, year FROM tracks ORDER BY cat_id asc";
$result = mysql_query ($requete,$db);

// count total number of appropriate listings:
$tcount = mysql_num_rows($result);

// count number of pages:
$tpages = ($tcount) ? ceil($tcount/$rpp) : 1; // total pages, last page number

$count = 0;
$i = ($page-1)*$rpp;
while(($count<$rpp) && ($i<$tcount)) {
    mysql_data_seek($result,$i);
    $query = mysql_fetch_array($result);

    // output each row:
    echo'<tr>
                        <td valign="top" width="112" height="27"><span class="font">'.$catno.'</span></td>
                        <td valign="top" width="536" height="27"><span class="font">'.$songname.'</span></td>
                        <td valign="top"><span class="font" height="27">'.$artistname.'</span></td>
                        <td valign="top" width="80" height="27"><span class="font">'.$ryear.'</span></td>
                    </tr>';

    $i++;
    $count++;
}

// call pagination function:
include("paginate.php");
echo paginate_three($reload, $page, $tpages, $adjacents);  

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Ok this time i managed to get it to display the results correctly but now i can only get the first page, when i go to next page it says not found.

 

Here is the code and the link:

http://www.nealeweb.com/steven2/

<?php

$rpp = 24; // results per page
$adjacents = 4;

$page = intval($_GET["page"]);
if($page<=0) $page = 1;

$reload = $_SERVER['PHP_SELF'];

$dbhost = "localhost";
$dbuser = "";
$dbpass = "";
$dbname = "";

// connect to your DB:
$link_id = mysql_connect($dbhost, $dbuser, $dbpass) or die(mysql_error());
mysql_select_db($dbname, $link_id);

// select appropriate results from DB:
$sql = "SELECT * FROM tracks ORDER BY cat_id asc";
$result = mysql_query($sql, $link_id);

// count total number of appropriate listings:
$tcount = mysql_num_rows($result);

// count number of pages:
$tpages = ($tcount) ? ceil($tcount/$rpp) : 3; // total pages, last page number

$count = 0;
$i = ($page-1)*$rpp;
while(($count<$rpp) && ($i<$tcount)) {
    mysql_data_seek($result,$i);
    $row = mysql_fetch_array($result);

$catno = $row['cat_id'];
$songname = $row['song'];
$artistname = $row['artist'];
$ryear = $row["year"];

    // output each row:
    echo'<tr>
					<td valign="top" width="112" height="27"><span class="font">'.$catno.'</span></td>
					<td valign="top" width="536" height="27"><span class="font">'.$songname.'</span></td>
					<td valign="top"><span class="font" height="27">'.$artistname.'</span></td>
					<td valign="top" width="80" height="27"><span class="font">'.$ryear.'</span></td>
				</tr>';

    $i++;
    $count++;
}
include("pagination3.php");
echo paginate_three($reload, $page, $tpages, $adjacents); 
?>

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I looked at those functions, and they do NOT build the link's right.  I would suggest using basic-pagination.  Then maybe you could go back and fix the link builds.

 

Hint:

Your current link it built like: http://www.nealeweb.com/steven2/index.php&page=3

However, it should be built like: http://www.nealeweb.com/steven2/index.php?page=3 <- it works on your server SEE

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