Question

I am trying to create a simple link. The issue is the file name is going to come from the database.

example: <a href="foldername/**FileNameInSQL.zip**">Download</a>

I have not worked much with MYSQL and have pieced together something that is working so far

<?php
$products_id = $_GET['id'];

mysql_connect("localhost", "username", "password") or die(mysql_error()); 

mysql_select_db("database") or die(mysql_error()); 

$sql = "select * from znc_product_extra_fields where products_id = '" . $products_id . "'"; 

$query = mysql_query($sql);

while ($row = mysql_fetch_array($query)) { 
echo $row['file_1']; 
}

?> 

When I run it it does just what I want, it echos the file name that is assigned to that specific row (item number)

But I am lost how to turn this into the link. The only thing I can think is somehow assigning this result to a variable and calling it while creating the link but I do not know how to take this result which is correct and actually use it! How would I take this filename and place in the link

<a href="foldername/**FileNameInSQL.zip**">Download</a>

Answer 1

PHP outputs whatever you want it to – text, HTML, XML, etc. So just output the HTML. I think what you want is:

echo "<a href=\"foldername/".$row['file_1']."\">Download</a>";

Although you shouldn’t be using the outdated mysql_* functions. Please see PDO (the best option) or mysqli.

To prevent SQL injection, use PDO::quote (if you are using PDO), or mysqli_real_escape_string (if you are using mysqli).