Question

我有一个搜索栏，用户可以搜索。所以当他们搜索时，他们会转到搜索页面，这将显示结果取决于他们搜索的内容，并且每页最多显示五个结果，所以如果有更多结果然后将显示分页链接。

但我的代码无效，请告诉我我的错误

这是我的 search.php代码

<?php
    include('include_Connection.php');
    if(isset($_GET["q"]))
         $q = $_GET['q'];

    $q=mysql_real_escape_string($_GET['q']);
    $res = mysql_query("SELECT COUNT(*) FROM scripts") or die($res."<br/><br/>".mysql_error());
    $get_total_rows = mysql_fetch_array($res) ; //total records

    //break total records into pages
    $pags = ceil($get_total_rows[0]/$item_per_page);    

    //create pagination
    if($pags > 1)
    {
        $pagination = '';
        $pagination .= '<ul class="pagination pagination-sm">';
        for($i = 1; $i<$pags; $i++)
        {
            $pagination .= '<li><a href="#" class="paginate_click" id="'.$i.'-page">'.$i.'</a></li>';
        }
        $pagination .= '</ul>';
    }

?>    

    <script type="text/javascript">
    $(document).ready(function() {
        $("#results").load("searchq.php", {'page':0}, function() {$("#1-page").addClass('active');});  //initial page number to load

        $(".paginate_click").click(function (e) {

            $("#results").prepend('<div class="loading-indication"><img src="ajax-loader.gif" /> Loading...</div>');

            var clicked_id = $(this).attr("id").split("-");
            var page_num = parseInt(clicked_id[0]); 

            $('.paginate_click').removeClass('active'); 

            $("#results").load("searchq.php", {'page':(page_num-1)}, function(){

            });

            $(this).addClass('active'); 

            return false;
        }); 
    });
    </script>

这是我的 searchq.php代码

<?php

include('include_Connection.php'); //include config file

//sanitize post value
$page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);

//validate page number is really numaric
if(!is_numeric($page_number)){die('Invalid page number!');}

//get current starting point of records
$position = ($page_number * $item_per_page);

//Limit our results within a specified range. 
$results = mysql_query("select Name_s,Price,Full_description,Img,ID_s,Pro_Status from scripts where Pro_Status = 'Active'  order by ID_s LIMIT $position, $item_per_page");

//output results from database
echo '<ul class="page_result">';
while($row = mysql_fetch_array($results,MYSQL_ASSOC))
{
    echo '<li id="item_'.$row["ID_s"].'">'.$row["ID_s"].'</li>';
}
echo '</ul>';
?>

Ajax和php搜索分页无法正常工作

0 个答案: