Passing variables with pagination - iterating through unique id per link

Posted on 16th Feb 2014 07:03 pm by admin

Hi everyone,
I have seen a few topics like this one. Still cant find the specific thing, so I am asking - does anyone know this?

I am trying to combine a mysql query with this code I found at php-mysql-tutorial.com.

Essentially this is the rub. The pages are working fine, but I am flumoxxed as to how to apply the $imgid to a page (viewing one image per page).

The imgid are in numerical order but do not begin at one, and there may be gaps - eg img 3, 5, 7,8 ,9 and 10 on a 6 page pagination.
I can pass the $page variable because it is a simple increment iteration, but how do I apply the changing $imgid value? so that pp1 has an imgid of 3 passed, pp2 is 5, pp3 is 7 etc. I expect it is something to do with combining a query and an array and applying it, but I'm having a brain-strain at that point.

Gratefully, here is my code stripped of excess ( I hope).

<?php

// back button depending on which gallery is parent
if(!isset($_REQUEST['random'])) {

$file= "gallery";

$varid = 0;
} else {
$file= "random";
$varid = 1;
}

echo "<fieldset><div id="back_but"><a href="gallery.php?blog=$file"> <br /></a></div>";

$url = "blog=picture&pics=picdetail&";

// how many rows to show per page
$rowsPerPage = 1;

// by default we show first page
$pageNum = 1;

// if $_GET['page'] defined, use it as page number
if(isset($_GET['page']))
{
$pageNum = $_GET['page'];
}

// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;


$query = "SELECT imgid, title, year, description, thumburl, imgurl, medium, style, clientname FROM images LEFT JOIN mediums ON images.mediumid = mediums.mediumid LEFT JOIN imgstyle ON images.styleid = imgstyle.styleid LEFT JOIN clients ON images.clientid = clients.clientid WHERE imgid = $imgid".
" LIMIT $offset, $rowsPerPage";

echo $query;
$result = mysql_query($query);
while($row=mysql_fetch_array($result, MYSQL_ASSOC))

{
$imgid = $row['imgid'];
$title = $row['title'];
$year = $row['year'];
$description = $row['description'];
$thumburl = $row['thumburl'];
$medium = $row['medium'];
$style = $row['style'];
$client = $row['clientname'];
}
?>

<fieldset><?php

// how many rows we have in database
$query = "SELECT COUNT(imgid) AS numrows FROM images WHERE tarot = $varid";
$result = mysql_query($query) or die('Error, query failed');
$row = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows'];

// how many pages we have when using paging?
$maxPage = ceil($numrows/$rowsPerPage);

// print the link to access each page
$self = $_SERVER['PHP_SELF'];
$nav = '';

for($page = 1; $page <= $maxPage; $page++)
{
if ($page == $pageNum)
{
$nav .= " $page "; // no need to create a link to current page
}
else
{

$nav .= " <a href="$self?$url&page=$page">$page</a> ";
}
}

// creating previous and next link
// plus the link to go straight to
// the first and last page

if ($pageNum > 1)
{
$page = $pageNum - 1;
$prev = " <a href="$self?$url&page=$page">[Prev]</a> ";

$first = " <a href="$self?$url&page=1">[First Page]</a> ";
}
else
{
$prev = '&nbsp;'; // we're on page one, don't print previous link
$first = '&nbsp;'; // nor the first page link
}

if ($pageNum < $maxPage)
{
$page = $pageNum + 1;
$next = " <a href="$self?$url&page=$page">[Next]</a> ";

$last = " <a href="$self?$url&page=$maxPage">[Last Page]</a> ";
}
else
{
$next = '&nbsp;'; // we're on the last page, don't print next link
$last = '&nbsp;'; // nor the last page link
}

// print the navigation link
echo $first . $prev . $nav . $next . $last;



?></fieldset>

Your Answer:

Login to answer

Other forums

Need help in log in and log out?
hi there all of u. i have recently created a site for someone. i have placed log in and log out and

Recode Abap Dynpro into Web Dynpro
Hi All,
A client has asked us to look at rearchitecting a custom transaction that was developed

MII Trends - add data onto chart object
Hello,

Can anyone please provide some thoughts on my current requirement:
its pretty

please help me... my password gets encrypted but not able to get in database
hai guys,
please help me... in the following function my password gets encrypted but no

How to show the difference between two data field in a database with php.
Hello php gurus,

how r u all... i'm not so well facing a typical problem please help me...<

browse folder only
hi,
is there a way i can browse or get the directory or folder only and not the file?
i just w

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
what is wrong with this

Code: [Select]<?php
if ($_SERVER['HTTP_REFERER'])
{

Redistributing dependent dlls
Hai all ,

I have created an application in VC++ using VS2008 in a development machine which r

I need help with formatting date from mysql! with php!
Hello guys

*Note: this is a php question not MySQL question, please do not move it to mysql s

Create multiple (n) arrays
Hi there,

I have the following need:

I have 2 arrays (coming from a databases)