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

same querie, or a new one??
Hi guys,

I'm still working on this drop down list. I've got the actual drop down list to wor

please hep to get values from a table row
can anyone tel me how I can select and get data from one row by pressing an Edit button in that row

admin with my register system?
Hey i wana make it so i can make a admin level on my register system , I'm kinda new to php to im no

understanding functions and classes
Code: [Select]
class person {
var $name = "Jimmy Goe";

function get_nam

Email Form Syntax Issue
I need the TO: in email to display To: CEO instead of To: abc@mail.com

How to alter the scri

help with multi-update
Now sure how to ask this really....
10g database if that matters.

I have a customer

PHP search multiple input field box help
I am having a problem with my search script. At current it will simply search by a selected date whi

script and html conflict in trying to create a header.
I have an error is occurring because of an html webpage with a "php require" at the top of

how do i make new line after *
First check this page here. and you see my report. Im pulling form a mysql db. I want to beable to m

Adding to an Int row in db
Hi, i have a database which houses all of the users of my site. One of the columns is for points whi