Pass image url from one page to another in PHP - php

Question

I am fetching data from database table and showing result in HTML table. What I want to achieve is that, there is image column in every row, and every image is fetched from its URL (stored in database). I am opening an image in new tab whenever it is clicked.
My question is that, how can I store url of image so that on the next page I don't want to show that real url?
The next page will look like this link:
www.example.com/full_size_image.php
and add
img tag
there to show that page
how can I store that specific image url when someone click on it?
<img src= "<?php echo $_SESSION['link'] ;">
My current code is :
while($result = $sql->fetch(PDO::FETCH_ASSOC))
{
echo "<tr>";
echo "<td>".$result['c']."</td>";
echo "<td>".$result['UserName']."</td>";
echo "<td>".$result['UserProblemKeyword']."</td>";
echo "<td> <a href ='".$result['UserProblemPicture']."' target='_blank'><img src='".$result['UserProblemPicture']."' height='62' width='62'> </a> </td>";
echo "</tr>";
}

Answer 1

Use a parameter via the GET Request?
So instead of directing to just /full_size_image.php direct to
/full_size_image.php?img=filename.png
then in your php code for /full_size_image.php
You can fetch the file name via
<?php
$img = isset($_GET['img']) ? htmlspecialchars($_GET['img']) : "";
if ($img == "") {
//img not set
}
//do whatever with the filename