Suggestions with pagination

Question 1

Question by dorin dorin

I’ve a problem with pagination. The php code of pagination query is:

$ <? if (isset($_GET["page"])) { 

    $Page = preg_replace("/[^0-9]/","", $_GET["page"]);
} else {
    $Page = 0;
}
$limit       = 10;
$StartFrom  = $limit * $Page;
$TotalFiles = mysql_num_rows(mysql_query("SELECT * FROM linkuri WHERE `categorie` = 'Afaceri' AND status = 1"));
$TotalPages = floor($TotalFiles / $limit); ?>

And code to display it:

$i = 0; while ($TotalPages >= $i) { echo '<a class="active imgf" style="opacity: 1;margin-bottom:3px; margin-top:3px;" href="afaceri.php?page='.$i.'">'.($i+1).'</a>';$i++;}

The problem is I am trying to make the display as: PAGES: "BACK" 1 2 3 4 5 "NEXT"

Answer

Answer by Starx

Starx

First I would like to suggest an improvement on your query. To count the total rows use:

SELECT count(*) as `totalpost` FROM linkuri WHERE `categorie` = 'Afaceri' AND status = 1

Then, you are create that pagination by doing something like this:

$totalPost = 50; //Dummy total post
$limit = 10;

$pages = $totalPost / $limit; //Giving 5

//Now we know there are five pages
for($i=1; $i<=$pages; $i++) {
  echo $i; // Better echo something like <a href="link">$i</a>
}

_{P.S: This is a very basic example}

After you get the hang of how to create the pagination effect, check this tutorial

How to paginate with PHP?

Question 2

Question by user366106

user366106

I want to create a pagination script using jquery UI’s slider widget. So far I have got the slider working but I dont know how to send request to the server to fetch new content.

So far this is my HTML

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8" />
    <title>jQuery UI Slider - Range slider</title>
    <link type="text/css" href="themes/base/jquery.ui.all.css" rel="stylesheet" />
    <script type="text/javascript" src="jquery.js"></script>
    <script type="text/javascript" src="jquery.ui.core.js"></script>
    <script type="text/javascript" src="jquery.ui.widget.js"></script>
    <script type="text/javascript" src="jquery.ui.mouse.js"></script>
    <script type="text/javascript" src="jquery.ui.slider.js"></script>      
    <style type="text/css">
        body { padding:5em; }
    </style>
    <script type="text/javascript">
    $(function() {
        $("#slider-range").slider({
            min: 1,
            max: 14,
            values: [1],
            slide: function(event, ui) {
                $(".values").html(ui.values[0];);
            }
        });
    });
    </script>
</head>
<body>
<div class="values" style="padding:2em;"></div>
<div id="slider-range"></div>

<div class="info" style="margin-top:2em; background:#CCC;padding:2em;">
Here is where the content will be displayed.
</div>

</body>

Thanks in Advance

Answer

Answer by Starx

Starx

Well, You can send a request at the slide event of your slider to send a request to the server at put the fetched data inside a div

  $("#slider-range").slider({
   min: 1,
   max: 14,
   values: [1],
   slide: function(event, ui) {
    var newPage = ui.values[0];
    $(".info").load("content.php", { page: newPage });
                                //load the content into a division
   }
  });

UPDATED
a sample of content.php

<?
     $recordsperpage = 15;
     if(!isset($_POST['page'] or empty($_POST['page']) { $page =1 ; }
     else { $page = $_POST['page']; }

     $limit = $page * $recordsperpage.",".$recordsperpage;
     $query = "SELECT * FROM yourtable LIMIT ".$limit;
     $result = mysql_query($query) or die(mysql_error());
     while($row = mysql_fetch_array($result)) {
          //Display the records in your pattern
     }
?>
?>

Question by dorin dorin

Answer by Starx

Question by user366106

Answer by Starx

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