Question

我尝试了一些搜索数据库表条目的代码。代码运行良好。我想在同一页面上显示div的结果。我无法获得这个的Ajax代码。我已经尝试了很多次。

形式

<head>
<script type="text/javascript" src="engine1/jquery.js"></script>
<script src="//code.jquery.com/jquery-1.10.2.min.js"></script>
<script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
<link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
<script  type="text/javascript">
    $(document).ready(function(){
       //can't get the ajax code
    });
</script>

</head>

<div class="login">
<form action="result.php" id="search_form" method="POST">

    <p><input name="query" autocomplete="off" id="list_search" type="search" required class="search" /></p>

    <p align="center">
        <input type="submit" id="click" name="click" class="but" value=" search"  />

    </p>
     </form>
  </div>

    <div class="coupons">

    </div>

result.php

if($_POST['click']) {
    $keyword = mysqli_real_escape_string($con,$_POST['query']); // always escape
    $keys = explode(" ", $keyword);

    $sql="SELECT  c.* , sc.* , sm.* ,ca.* from store_category sc INNER JOIN store_manufacture sm ON sm.sm_id=sc.store_id INNER JOIN categories ca ON ca.cat_id=sc.cat_id INNER JOIN coupons c on c.c_sc_id=sc.sc_id WHERE c.c_name LIKE '%$keyword%' OR sm.sm_brand_name LIKE '%$keyword%' OR ca.cat_name LIKE '%$keyword%' OR c.c_description LIKE '%$keyword%' LIMIT 10";

    foreach ($keys as $k) {
        $sql.="OR c.c_name LIKE '%$k%' OR sm.sm_brand_name LIKE '%$k%' OR ca.cat_name LIKE '%$k%' OR c.c_description LIKE '%$k%'";
    }

    $result = mysqli_query($con, $sql);
    while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
        echo $row['c_name'];
        echo $row ['c_description'];
        echo $row['sm_brand_name'];
    }
}

搜索结果显示在div的同一页面上

0 个答案: