Pagination Loader without Database (for pictures)

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JanMolendijk
Posts: 140
Location: Holland Rotterdam

Pagination Loader without Database (for pictures)

Dear ADmin i hope you can help me with my next problem

I have for the members seperated folders for pictures...
But some members have to mutch pictures to load in once

I`m searching for 7 hours for a Pagination Loader without Database for the pictures

Code: Select all

<?php
		  
$files = glob("../images/$users[email]/*.*");

for ($i=0; $i<count($files); $i++)
{
	$num = $files[$i];

	echo '<div class="col-xs-6 col-md-3">
	<div class="thumbnail">
			<a href="'.$num.'" rel="shadowbox"><img src="'.$num.'" class="img-responsive" > </a>
			<br>
<link href="ratingfiles/ratings.css" rel="stylesheet" type="text/css" />

<div class="srtgs" id="rt_'.$num.'"></div>
<script src="ratingfiles/ratings.js" type="text/javascript"></script>
			<div class="caption">
			<p></div>
			
			</div>
			</div>';
}
?>	  
</div>		

Admin
Hello,
You can use the Pagination Class from this page:
http://coursesweb.net/php-mysql/paginat ... -script_s2

- It can be used to paginate data from database, items of an array, and even a large content string.
- See the documentation and examples in that page.

Basically, you have to include the class, and to add the array with pictures files to the getArrRows($files ); method.
- In the class code, in the public function getArrRows($arr) method define the html you want to render:

Code: Select all

// HERE ADDS HTML TAGS TO FORMAT THE ZONE THAT CONTAINS EACH ELEMENT
for($i=0; $i<$nre; $i++) {
  $re_cnt .= '<div class="content">'. $ar_page[$i]. '</div>';
}
- Something like this:

Code: Select all

// HERE ADDS HTML TAGS TO FORMAT THE ZONE THAT CONTAINS EACH ELEMENT
for($i=0; $i<$nre; $i++) {
  $re_cnt .='<div class="col-xs-6 col-md-3"><div class="thumbnail">
<a href="'. $ar_page[$i]. '" rel="shadowbox"><img src="'. $ar_page[$i]. '" class="img-responsive" > </a><br>
<div class="srtgs" id="rt_'. $ar_page[$i]. '"></div>
<div class="caption"></div>
</div></div>';
}
$re_cnt .='<link href="ratingfiles/ratings.css" rel="stylesheet" type="text/css" />
<script src="ratingfiles/ratings.js" type="text/javascript"></script>';

return $re_cnt;

JanMolendijk
Dear Admin In the class code, in the public function getArrRows($arr)
i changed the code like you said

I don`t get it i have the follow code placed into the test_array
but i cant get $id into echo $objPg->getLinks();

the $id i need for the next pagination

Code: Select all

<?php 
include('../connection.php');
$id = (int) $_GET['id'];
$sql=mysqli_query($conn,"SELECT * FROM `user` WHERE id='" . $id . "' ");
$users=mysqli_fetch_assoc($sql);
 ?>

Code: Select all

<?php 
$files = glob("../../images/$users[email]/*.*");
for ($i=0; $i<count($files); $i++)
{
$num = $files[$i];
}
 ?>

Code: Select all

<?php 
// Array with data
$source = array('a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','y','x','z');

// include the Pagination class
include('class.pagination.php');

// create object instance of the class
$objPg = new Pagination();

// if you want to change the number of rows /elements added in page, set the "rowsperpage" property
// $objPg->rowsperpage = 8;

// output /display the conntent
echo $objPg->getArrRows($files);

// output / display pagination links
echo $objPg->getLinks();

?>

Admin
I updated the class code.
Get again the class from the link added in the previous response, then use this syntax to add an $id in the pagination links:

Code: Select all

$str ='id='.$id;
echo $objPg->getLinks($str);

JanMolendijk
Thanks for all support again &nd again

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