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.
Did you know?Explore Trending and Topic pages for more stories like this.

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>
No comments posted yet

Your Answer:

Login to answer
184 Like 30 Dislike
Previous forums Next forums
Other forums

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

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

Timer control causing error
I recently decided to add a timer control to an existing page that uses AJAX on my site. As soon as

Onclick problem in Firefox
Hi,
I am using a . It doesn't seem to

Saving PHP output as a file
I wondered if anyone knows how to save the rendered output of a PHP script as a file?

I have

Sessions Value Not Saved
Hi,

Im making a login form and im using this code:
Code: if($login=="true"){

Checking if field is empty AND using !is_numeric
Hi lads,

Need to be able to display an error message (using isset?) that says "Cant lea

why does my session end?
my connect.php starts the session just so you know
i can navigate arround my site fine except whe

HOW to get the bind variables list.
I've the following problem : I've some SQL queries stored in my DB as VARCHAR2 values.
I need t

error help - Dynamic Image
I've been working on making my site less cluttered in the directories and more secure lately. In an

Performance impact of cookies
Hi, I was just wondering what impact there would be in terms of performance if you where to set then

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash