使用 slug url 从 mysql 数据库中获取文章

Question

我正在开发自己的 PHP Mysql 网站。

我曾经像这样通过 id 获取数据：mysite.com/articles.php?id=12

现在我想用 slug 更改 url：

mysite.com/articles/google-search

或

mysite.com/articles.php?article=google-search

我不想用id和numbers

我的table:

+----+---------------+---------+------------------------------------+
| id | title         | article |   urlslug                          |
+----+---------------+---------+------------------------------------+
| 12 | google search | xxxxxxx |   google-search                    |
| 13 | bing yahoo    | xxxxxxx |   bing-yahoo                       |
| 14 | friendly seo  | xxxxxxx |   friendly-seo                     |
+-------------------------------------------------------------------+

我使用下面的代码通过 id 获取数据：

$id = $_GET['id'];
$id = mysqli_real_escape_string($conn,$id);
$query = "SELECT * FROM `table` WHERE `id`='" . $id . "'";
$result = mysqli_query($conn,$query);

while($row = mysqli_fetch_array($result)) {
echo ($row['title']);
echo ($row['article']);    }

我通过替换 urlslug 来尝试上面的代码，它说 Invalid ID specified. 我用谷歌搜索甚至搜索堆栈问题我没有得到任何帮助。请帮助我。提前致谢。

Answer 1

如果url是这样的mysite.com/articles.php?article=google-search 然后从 url 获取 article 而不是 id 并将条件更改为 urlslug 而不是 id.

$slug = $_GET['article'];
$slug = mysqli_real_escape_string($conn,$slug);
$query = "SELECT * FROM `table` WHERE `urlslug`='" . $slug. "'";
$result = mysqli_query($conn,$query);

 //Since  slug is unique you will get only 1 result so no need to loop

 $row = mysqli_fetch_array($result);
 echo $row['title'];
 echo $row['article'];

Answer 2

您可以在 Root/.htaccess 中使用以下规则

RewriteEngine On


RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-f
RewriteRule ^articles/([^/]+)/?$ /articles.php?id= [NC,L]

这将重写

example.com/articles/123

至

example.com/articles.php?id=123

使用 slug url 从 mysql 数据库中获取文章

fetching article from mysql db with slug url

php

mysql

.htaccess

slug