Question

我在上一个（第三个）动态下拉列表中遇到问题 - 结果错误。

<?php
include('dbconfig.php');
$action = $_REQUEST['action'];
if ($action=='showAll')   
{$id = "0";}
else $id=$_POST['id'];

switch($action)
{
case 'showAll';
    $stmt=$DB_con->prepare('SELECT * FROM slist GROUP BY site');
    $stmt->execute();
break;
case 'site';
    $stmt=$DB_con->prepare('SELECT menu FROM slist WHERE site=:id GROUP BY menu ORDER BY menu');
    $stmt->execute(array(':id'=>$id));
break;
case 'menu';
    $stmt=$DB_con->prepare('SELECT categ FROM slist WHERE menu=:id GROUP BY categ ORDER BY categ');
    $stmt->execute(array(':id'=>$id));
break;
case 'categ';
    $stmt=$DB_con->prepare('SELECT links FROM slist WHERE categ=:id GROUP BY categ ORDER BY links');
    $stmt->execute(array(':id'=>$id));
break;
default: 
    $stmt=$DB_con->prepare('SELECT * FROM slist GROUP BY site');
    $stmt->execute();
break;
}
?>

SELECT links FROM slist WHERE categ=:id GROUP BY categ ORDER BY links $action=="categ"返回错误的结束值（从列表中的第一个结果中获取结果），因为查询应如下所示：SELECT links FROM slist WHERE menu=:id AND categ=:id GROUP BY categ ORDER BY links。

如何将“menu”变量的值传递给此查询？

SQL表列名和表中的几条记录：

`id`, `site`, `menu`, `categ`, `links`
28, 'CoursesWeb.net', 'PHP-MySQL', 'Tutorials', 'http://coursesweb.net/php-mysql/tutorials_t'
34, 'CoursesWeb.net', 'JavaScript', 'Tutorials', 'http://coursesweb.net/javascript/tutorials_t'
46, 'CoursesWeb.net', 'Flash-AS3', 'Tutorials', 'http://coursesweb.net/flash/tutorials_t'

如何将少数sql选择转换为多个选择查询

0 个答案: