Pagination Loader without Database (for pictures)

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JanMolendijk
Posts: 107
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

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<?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:

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// 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:

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// 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

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

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

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<?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:

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

JanMolendijk
Thanks for all support again &nd again

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