php - 如何使sql查询动态化？

Question

我正在使用这个pagination class并正在寻找一种方法使 sql 查询更加动态，而不是对其进行硬编码。

我有一个 3 <li>我希望成为过滤按钮的元素，这意味着当用户单击这些元素之一时，我希望它发送 id，以便我可以在 sql 查询中使用它。

所以对于 $sql = "select * from explore where category='marketing'"; (如下所示)。当用户单击“汽车”按钮时，会将上面的类别更改为汽车。

对此的任何帮助将不胜感激，谢谢。

这就是我的主页:

<script type="text/javascript"
    src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script>
<script type="text/javascript" src="jquery_page.js"></script>


<?php
    //Include the PS_Pagination class
    include('ps_pagination.php');
    //Connect to mysql db
    $conn = mysql_connect('localhost', 'root', 'root');
    mysql_select_db('ajax_demo',$conn);
    $sql = "select * from explore where category='marketing'";
    //Create a PS_Pagination object
    $pager = new PS_Pagination($conn, $sql, 3, 11, 'param1=valu1&param2=value2');
    //The paginate() function returns a mysql
    //result set for the current page
    $rs = $pager->paginate();
    //Loop through the result set
    while($row = mysql_fetch_assoc($rs)) {
            echo "<table width='800px'>";

                echo "<tr>";
                    echo"<td>";
                    echo $row['id'];
                    echo"</td>";

                    echo"<td>";
                    echo $row['site_description'];
                    echo"</td>";

                    echo"<td>";
                    echo $row['site_price'];
                    echo"</td>";
                echo "</tr>";

            echo "</table>";
    }

        echo "<ul id='pagination'>";

            echo "<li>";
            //Display the navigation
            echo $pager->renderFullNav();
            echo "</li>";

        echo "</ul>";

    echo "<ul id='filter'>";

        echo "<li id='marketing'>";
            echo "Marketing";
        echo "</li>";

        echo "<li id='automotive'>";
            echo "Automotive";
        echo "</li>";

        echo "<li id='sports'>";
            echo "Sports";
        echo "</li>";

    echo "</ul>";
?>

Answer 1

看起来像这一行

$pager = new PS_Pagination($conn, $sql, 8, 3, 'param1=valu1&param2=value2');

应该编辑以在运行sql查询之前携带要测试的变量....可能类似于:

$pager = new PS_Pagination($conn, $sql, 8, 3, 'param1=automotive&param2=sports');

那就试试这个

if($_POST["param1"]=="automotive")
{
$sql = "select * from explore where category='automotive'";
}

并尝试解决它，看看是否可以获得您想要的结果。我以前从未使用过该类，但看起来您必须使用这些参数才能获得所需的结果