Error in jquery paging

Error in jquery paging

Hi,

my application has clientside paging.

following is working code .It contains 10 records and display only four.

paging also contains forward and backward button .


  1. <script type="text/javascript" language="javascript">

        $(document).ready(function () {
            $('#pages').empty();
            $('#pages').show();
            var cnt = 10;
            var $rcount = "There are " + cnt + " Records";
            //alert($rcount);
            $('#ContentPlaceHolder1_hdint').val(1).val();
            var disp = $('<div id="dp"></div>');
            $(disp).html($rcount).css('color', 'Green');
            var $pagenumbers = $('<div id="pages"></div>');
 var html = '<a id="prev" title="previous" href=\"#\"> <<</a>';
var k = cnt / 3;
       for (i = 0; i < k; i++) {
   html = html + '<a href="#" id="' + (i + 1) + '" class="page">' + (i + 1) + '</a>';
            }

 html = html + '<a title="Next" id="next" href=\"#\"> >></a>';
       $(html).appendTo($pagenumbers);
     $pagenumbers.insertAfter('#st');
    $(disp).insertAfter($pagenumbers);

});
</script>




in above code prev and next are previous and next buttons.

for moving forward direction and display next 4 records, i try following example but showing error

  1. $('#next').click(function () {
                $('#pages').empty();
                               
               
          $('#ContentPlaceHolder1_hdint').val(1).val();
             var disp = $('<div id="dp"></div>');
          var $pagenumbers = $('<div id="pages"></div>');
var html = '<a id="prev" title="previous" href=\"#\"> <<</a>';
                i += 3;
                k += 3;
                $('#pages').empty();
                
                for (i = 0; i < k; i++) {
html = html + '<a href="#" id="' + (i + 1) + '" class="page">' + (i + 1) + '</a>';
                }

html = html + '<a title="Next" id="next" href=\"#\"> >></a>';
                $(html).appendTo($pagenumbers);
                $pagenumbers.insertAfter('#st');
                $(disp).insertAfter($pagenumbers);
            });





error is it showing 7 records.

i need only 5,6, and 7.

how to solve this.

Regards

Baiju