Solved

PHP pagnation

Posted on 2010-08-21
3
341 Views
Last Modified: 2013-12-13
Hello,

I've got the following.
<div class="entry"> 
					<table class="browsetable"> 
						
						<?php 
						$getartists_sql = mysql_query("SELECT artistID, artistname FROM artists WHERE artistname LIKE '".mysql_real_escape_string($browse)."%'");
						$color = 1;
						while($row_browse = mysql_fetch_array($getartists_sql, MYSQL_ASSOC))
						{
							if($color == 1) {
							?>
							<tr> 
								<td class="even"><a href="#" title=""><?php echo stripslashes($row_browse['artistname']); ?></a></td> 
								<td class="even lyriccount">123 lyrics</td> 
							</tr> 								
							<?php 
							$color = 2;
							}
							else{
							?>
							<tr> 
								<td class="uneven"><a href="#" title=""><?php echo stripslashes($row_browse['artistname']); ?></a></td> 
								<td class="uneven lyriccount">123 lyrics</td> 
							</tr> 
							<?php
							$color = 1;
							}
						}
						?>
					</table> 	
				</div> 

Open in new window


and it returns a lot of artistnames, but now I want pagnation.
So when there are more than 20 names, make a new page with link.
more than 40 names, 2 pages with link etc.
0
Comment
Question by:Serellyn
3 Comments
 
LVL 5

Assisted Solution

by:ploftin
ploftin earned 150 total points
ID: 33493505
Something like this:
<?php
$currentPage = $_SERVER["PHP_SELF"];

$maxRows_Recordset1 = 20;
$pageNum_Recordset1 = 0;
if (isset($_GET['pageNum_Recordset1'])) {
  $pageNum_Recordset1 = $_GET['pageNum_Recordset1'];
}
$startRow_Recordset1 = $pageNum_Recordset1 * $maxRows_Recordset1;

mysql_select_db($database_Music, $Music);
$query_Recordset1 = "SELECT * FROM Artists";
$query_limit_Recordset1 = sprintf("%s LIMIT %d, %d", $query_Recordset1, $startRow_Recordset1, $maxRows_Recordset1);
$Recordset1 = mysql_query($query_limit_Recordset1, $Music) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);

if (isset($_GET['totalRows_Recordset1'])) {
  $totalRows_Recordset1 = $_GET['totalRows_Recordset1'];
} else {
  $all_Recordset1 = mysql_query($query_Recordset1);
  $totalRows_Recordset1 = mysql_num_rows($all_Recordset1);
}
$totalPages_Recordset1 = ceil($totalRows_Recordset1/$maxRows_Recordset1)-1;

$queryString_Recordset1 = "";
if (!empty($_SERVER['QUERY_STRING'])) {
  $params = explode("&", $_SERVER['QUERY_STRING']);
  $newParams = array();
  foreach ($params as $param) {
    if (stristr($param, "pageNum_Recordset1") == false && 
        stristr($param, "totalRows_Recordset1") == false) {
      array_push($newParams, $param);
    }
  }
  if (count($newParams) != 0) {
    $queryString_Recordset1 = "&" . htmlentities(implode("&", $newParams));
  }
}
$queryString_Recordset1 = sprintf("&totalRows_Recordset1=%d%s", $totalRows_Recordset1, $queryString_Recordset1);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>

<body>
<div class="entry"> 
  <table class="browsetable"> 
						
						<?php 
						$color = 1;
						do {
							if($color == 1) {
							?>
							<tr> 
								<td class="even"><a href="#" title=""><?php echo stripslashes($row_Recordset1['artistname']); ?></a></td> 
								<td class="even lyriccount">123 lyrics</td> 
							</tr> 								
							<?php 
							$color = 2;
							}
							else{
							?>
							<tr> 
								<td class="uneven"><a href="#" title=""><?php echo stripslashes($row_Recordset1['artistname']); ?></a></td> 
								<td class="uneven lyriccount">123 lyrics</td> 
							</tr> 
							<?php
							$color = 1;
							}
						} while ($row_Recordset1 = mysql_fetch_assoc($Recordset1));
						?>
					</table>
                    <p>&nbsp;<a href="<?php printf("%s?pageNum_Recordset1=%d%s", $currentPage, 0, $queryString_Recordset1); ?>">First</a> <a href="<?php printf("%s?pageNum_Recordset1=%d%s", $currentPage, max(0, $pageNum_Recordset1 - 1), $queryString_Recordset1); ?>">Previous</a> <a href="<?php printf("%s?pageNum_Recordset1=%d%s", $currentPage, min($totalPages_Recordset1, $pageNum_Recordset1 + 1), $queryString_Recordset1); ?>">Next</a> <a href="<?php printf("%s?pageNum_Recordset1=%d%s", $currentPage, $totalPages_Recordset1, $queryString_Recordset1); ?>">Last</a></p>
</div>
</body>
</html>
<?php
mysql_free_result($Recordset1);
?>

Open in new window

0
 
LVL 11

Accepted Solution

by:
Rajesh Dalmia earned 200 total points
ID: 33494130
you can use this freee php pagination class http://phpsense.com/php/php-pagination-script.html
it's easy to implement
0
 
LVL 14

Assisted Solution

by:john-formby
john-formby earned 150 total points
ID: 33494295
Hi,

Please check the following script: http://phpfreak.co.uk/comments.php?bID=17

It is a fully commented example which is easy to implement.

Hope this helps,

John
0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Author Note: Since this E-E article was originally written, years ago, formal testing has come into common use in the world of PHP.  PHPUnit (http://en.wikipedia.org/wiki/PHPUnit) and similar technologies have enjoyed wide adoption, making it possib…
Nothing in an HTTP request can be trusted, including HTTP headers and form data.  A form token is a tool that can be used to guard against request forgeries (CSRF).  This article shows an improved approach to form tokens, making it more difficult to…
The viewer will learn how to dynamically set the form action using jQuery.
The viewer will learn how to create a basic form using some HTML5 and PHP for later processing. Set up your basic HTML file. Open your form tag and set the method and action attributes.: (CODE) Set up your first few inputs one for the name and …

685 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question