如何在PHP中创建mySQL数据库及其相应文件（超链接）的文件名HTML表

Question

I have a form that submits a file to the server in the folder uploads and also saves the file name(as stored in the server) as well as other file details in my database. Now what I'm trying to do, is create a HTML table that displays file details as well as the corresponding file(In hyperlink of course).

Is there any way to this? An idea I had is to link the file name to the corresponding file in my uploads/ directory but I have no idea on how to achieve that.

Here is my table uploaded_content where the form data is stored in my database

================================================
id |Description | filename                     |
------------------------------------------------
1  |Information | 3223-2323-4334-32-slajjjq.txt|
2  |Users       | 3223-2323-4344-33-slik.txt   |
================================================

And here is an example of how my uploads/ directory looks like (Notice that their names are saved in my database)

3223-2323-4334-32-slajjjq.txt
3223-2323-4344-33-slik.txt

So now I want to create a HTML table with the following columns

id |Description | filename  |File   

Answer 1

It is not the best way to do it, but i just wanna show you how to do that.

Step 1 - Create mysql DB.

CREATE TABLE `uploaded_content` (
  `id` int(11) NOT NULL,
  `description` text NOT NULL,
  `filename` text NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

Step 2 - Insert data to this DB.

INSERT INTO `uploaded_content` (`id`, `description`, `filename`) VALUES
(1, 'Information', '3223-2323-4334-32-slajjjq.txt'),
(2, 'Users', '3223-2323-4344-33-slik.txt');

Step 3 - Make config.php:

<?php
    $conn = new PDO('mysql:host=localhost;dbname=stack', 'root', '');
    $q = $conn->query("SELECT * FROM uploaded_content WHERE id");
    $f = $q->fetchAll();
?>

Step 4 - Create index.php with this code:

<?php
include('config.php');
?>

<table>
    <tr>
      <th>ID</th>
      <th>Description</th> 
      <th>File</th>
      <th>Download link</th>
    </tr>
<?php 
foreach ($f as $g) :?>
  <tr>
    <td> <?php echo $g['id']; ?></td>
    <td> <?php echo $g['description']; ?></td>
    <td> <?php echo $g['filename']; ?></td>
    <td><a href="https://example.com/uploads/<?php echo $g['filename']; ?>">Download</a></td>
  </tr>
<?php endforeach;?>
</table>

When you finish that, you get html page like this:

<table>
    <tr>
      <th>Id</th>
      <th>Description</th> 
      <th>File</th>
      <th>Download link</th>
    </tr>
    <tr>
      <td>1</td>
      <td>Information</td>
      <td>3223-2323-4334-32-slajjjq.txt</td>
      <td><a href="https://example.com/uploads/3223-2323-4334-32-slajjjq.txt">Download</a></td>
    </tr>
    <tr>
      <td>2</td>
      <td>Users</td>
      <td>3223-2323-4344-33-slik.txt</td>
      <td><a href="https://example.com/uploads/3223-2323-4344-33-slik.txt">Download</a></td>
    </tr>
</table>

</div>