PHP display image from link - php

Question

I am trying to get my image to display when clicked via the link. However when I click on the link it finds the image id as displayed in the url link ok but does not display any image. Can you please help?
<?php
//sets up thisPage
$pageSize=10;
if (isset($_POST["thisPage"])) $thisPage=$_POST["thisPage"];
else $thisPage=1;
//selects all distinct expenses that have been uploaded
$dbQuery="SELECT * FROM images WHERE user_id = '$userID' ";
$dbResult=mysqli_query($db_connection, $dbQuery) or die(mysqli_error($db_connection));
echo "<table cellspacing=\"5\" class=\"recordsTableBG\"> <thead
class=\"recordsTableHeader\">";
echo '<tr> <th>ID</th><th>Amount</th><th>Description</th><th>Filename</th>
<th>Project ID</th><th>Status</th></tr></thead>';
echo '<tr class="alternateRowColor">';
'<tr>';
while ($dbRow=mysqli_fetch_array($dbResult)){
echo "<img src = 'uploaded/$image' width = '200' height = '200'>";
// display row with expense
echo '<td>'. $dbRow['id'] .'</td>';
echo '<td>'. $dbRow['user_id']. '</td>';
echo '<td><a href='.$_SERVER['PHP_SELF'].'?imageid='.$dbRow['id'].'>
Click here to view image</a></td>';
}
echo "</table>";
echo "</form>";
?>
<!-- add submitt button
close form -->
</div>

Answer 1

You have a mixture of two approaches going on here, and that's where the confusion lies.
First, notice your
<img src='uploaded/$image' width = '200' height = '200'>
The browser is going to see something like
<img src='uploaded/picture_of_cat.jpg' width='200' height='200'>
and render a 200px by 200px image of a cat. I'm assuming that's your thumbnail. (As an aside, I notice that it's not enclosed in <td></td> even though it's in the table row).
In another part of the table, you have
<a href='.$_SERVER['PHP_SELF'].'?imageid='.$dbRow['id'].'>Click here to view image</a>
which is going to render as something like
<a href='http://example.com/index.php?imageid=1234'>Click here to view image</a>
When the user clicks that link, it's going to make a GET request to the server with imageid equal to 1234. What is the server going to do with that imageid? In the code you've posted, nothing.
You have two choices:
if you want to keep the link as it is, you'll have to write some code that will take the imageid value, find the appropriate image, and return it to the browser as image data - that means setting the appropriate headers and sending the data back as binary data.
the simpler way to do it would be to replace the URL in the link with the same one you have in your <img> tag - when the user clicks on it, the server will just return the image.