Question

我有一个页面显示数据库中文章的标题。它们是链接，点击后转到一个单独的页面，我想根据ID显示标题和内容。链接如下所示：

        echo '<h2><a href="../../pages/content_page/contents_page.php?id='.$review_id.'">'.$title.'</a></h2>';

，显示它们的页面包含以下代码：

<?php
while ($row = mysql_fetch_assoc($query)){
    $review_id = $row['review_id'];
    $title = $row['title'];
    $review = $row['review'];
} ?>
<h1><?php echo $title; ?></h1><br /><br />

<p><?php echo $review; ?></p>

但是当我点击链接时，网址会显示正确的ID号，但是我点击的每个链接显示的标题和内容都是相同的。如何让它们针对每个单独的ID进行更改？提前谢谢！

编辑：

完整代码：

<?php

include '../../mysql_server/connect_to_mysql.php';

$query = mysql_query("SELECT `review_id`, `title`, `review` FROM `reviews`");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Posts!</title>

<body>

<?php
while ($row = mysql_fetch_assoc($query)){
    $review_id = $row['review_id'];
    $title = $row['title'];
    $review = $row['review'];
} ?>
<h1><?php echo $title; ?></h1><br /><br />

<p><?php echo $review; ?></p>
</body>
</html>

Answer 1

您正在提取所有行。您需要WHERE子句。

$review_id = intval($_GET['review_id']);
$resource = mysql_query("SELECT title, review FROM reviews
                         WHERE review_id = $review_id");
if (!$resource) {
    trigger_error(mysql_error());
}
# etc..

Answer 2

尝试这样的事情......

<?php
$id = $_GET['id'];
$query = mysql_query("SELECT `title`, `review` FROM `yourtable` WHERE `review_id` = '$id'");

$row = mysql_fetch_assoc($query);
//$review_id = $row['review_id']; **You will not need this because you already have it in the URL
$title = $row['title'];
$review = $row['review'];
?>

<h1><?=$title?></h1><br /><br />
<p><?=$review?></p>

编辑：更改：

$query = mysql_query("SELECT `review_id`, `title`, `review` FROM `reviews`");

要：

$query = mysql_query("SELECT `title`, `review` FROM `reviews` WHERE `review_id` = '".$_GET['id']."'");

另外......您需要对$_GET['id']进行一些过滤，例如：

preg_match('/^[0-9]{1,9}$/', $_GET['id'])

（假设您的ID是数字）

如何链接到数据库页面

2 个答案: