Question

i am using jquery ajax pagination code, i need some changes.

<?php
$per_page = 4;
$sql = "select * from portfolio  ";
$rsd = mysql_query($sql);
$count = mysql_num_rows($rsd);
$pages = ceil($count/$per_page)
?> 

the above code calculates the number of pages then

  <?php
    //Show page links
    for($i=1; $i<=$pages; $i++)
    {
        echo '<li  class="pagenum" id="'.$i.'">'.$i.'</li>';
    }
?>

this code display pagination button that is basically

<li class="pagenum" id="'.$i.'">'.$i.'</li>

all the problem is here with $i .

if $per_page=4 then 7 pagination button appears .

pagination occur on basis of

id="'.$i.'"

this way

$("#paging_button li").click(function(){
        //show the loading bar
        showLoader();

        $("#paging_button li").css({'background-color' : ''});
        $(this).css({'background-color' : '#ccc'});

        $("#contentt").load("data.php?page=" + this.id, hideLoader);
    });

what i want is to display only next and previous button . instead of seven pagination buttons .

need to change to logic/code .

please help

thanks

Answer 1

    var pager = $('<div id="pager" class="pagination"></div>');
    // adds the controls of pagination
    pager.pagination(maxentries, {
        items_per_page : limit,
        next_text : '>',
        prev_text : '<',
        num_display_entries : 5,
        load_first_page : false,
        callback : callback
    });