使用slug url从mysql db中获取文章

Question

Im developing my own PHP Mysql site.

I used to fetch data by id like this : mysite.com/articles.php?id=12

now i want to change the url with slug:

mysite.com/articles/google-search

or

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

I dont want to use id and numbers.

My table:

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

I used the below code to get data by id:

$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']);    }

I tried the above code by substituting urlslug it says Invalid ID specified. I googled and even searched in stack questions i didnt get any help. Please assist me. Thanks in advance.

Answer 1

If url is like this mysite.com/articles.php?article=google-search then instead of id get the article from url and change the condition to urlslug instead of 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'];