如何在此分页中添加过滤器按钮?

时间:2010-04-26 02:06:25

标签: php pagination

嘿,我想添加一个按钮(链接),单击该按钮将过滤分页结果。

我是php(和一般编程)的新手,想添加一个像'Automotive'这样的按钮,点击它时会更新我的分页脚本中的2个mysql查询,如下所示:

正如您所看到的,汽车类别是硬编码的,我希望它是动态的,因此当点击链接时,它会将查询的类别部分中的id或类放置。

1:

$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'"));

2:

$get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page");

这是我正在使用的整个当前的php分页脚本:

<?php

    //connecting to the database
    $error = "Could not connect to the database";
    mysql_connect('localhost','root','root') or die($error);
    mysql_select_db('ajax_demo') or die($error);

    //max displayed per page
    $per_page = 2;

    //get start variable
    $start = $_GET['start'];

    //count records
    $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'"));

    //count max pages
    $max_pages = $record_count / $per_page; //may come out as decimal

    if (!$start)
       $start = 0;

    //display data
    $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page");
    while ($row = mysql_fetch_assoc($get))
    {
     // get data
     $name = $row['id'];
     $age = $row['site_name'];

     echo $name." (".$age.")<br />";

    }

    //setup prev and next variables
    $prev = $start - $per_page;
    $next = $start + $per_page;

    //show prev button
    if (!($start<=0))
           echo "<a href='pagi_test.php?start=$prev'>Prev</a> ";

    //show page numbers

    //set variable for first page
    $i=1;

    for ($x=0;$x<$record_count;$x=$x+$per_page)
    {
     if ($start!=$x)
        echo " <a href='pagi_test.php?start=$x'>$i</a> ";
     else
        echo " <a href='pagi_test.php?start=$x'><b>$i</b></a> ";
     $i++;
    }

    //show next button
    if (!($start>=$record_count-$per_page))
           echo " <a href='pagi_test.php?start=$next'>Next</a>";

    ?>

2 个答案:

答案 0 :(得分:0)

包含2个链接/类别的示例:

<a href='script.php?category=automotive'>automotive</a> <a href='script.php?category=sports'>sports</a>

Inside script.php:

$category = mysql_real_escape_string($_GET['category']);
$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='$category'"));
...
$get = mysql_query("SELECT * FROM explore WHERE category='$category' LIMIT $start, $per_page");

答案 1 :(得分:0)

编辑:我只是发布了这个答案,以回应OP说他不熟悉编程。良好的卫生习惯在一开始就很容易学习,成为习惯,或者很晚以后很难养成习惯。

http://us2.php.net/manual/en/security.database.sql-injection.php

另请阅读有关数据库安全性的文章。将查询编写为:

会有很大改进
$start = mysql_real_escape_string($_GET['start']);
settype($start, 'integer');
$category = mysql_real_escape_string($_GET['category']);
$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='%s'", $category));
...
$get = mysql_query("SELECT * FROM explore WHERE category='%s' LIMIT %d, %d", $category, $start, $per_page);

编写额外的安全代码总是值得的,即使只是为了练习。数据库的基本安全规则是:永远不要信任用户输入,始终检查生成的输出。由于输入来自查询字符串并且针对数据库运行,因此必须对其进行过滤。