I have been searching for help from various forums and similar posts, but without any progress. I have three pages, one that lets me insert information about projects into my database, a second that show the images and names of every project in the database, and a third page which I want to have a function that shows the image and name of the selected project in the second page.
Code on the second page(dashboardadmin.php):
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
$conn = mysqli_connect("localhost","root","","wildfire");
if(mysqli_connect_errno())
{
echo mysqli_connect_error();
}
$sql= "SELECT pid, project_name, image, image_type FROM project";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
echo "<form action='omprojekt.php' method='post'>
<div id='comp' name='comp'>
<img src=pic.php?pid=".$row['pid']." width=100xp height=100xp/>"." ".$row['project_name']."
</div>
<input type='submit' name='submit' value='Choose' />
</form>";
}
}
else {
echo "0 results";
}
mysqli_close($conn);
?>
Code on the third page (omprojekt.php):
<?php
/* Tried both of the $val variables but of course only one at a time. This is only to show you what I have tried. */
$val = isset($_POST['comp']) ? $_POST['comp'] : '';
$val = $_POST['comp'];
if(isset($_POST['submit'])){
echo "$val";
}
?>
In the last code you can see that I have two $val
variables, but I have only used one of them at a time in my codes. The purpose of showing both of them here is to show you that I have tried both of them.
What I want to do is to make the third page show the image and name of the selected project in the second page. As you see, I have tried to retrieve the content from the DIV using the same "name". The problem is that the third page(omprojekt.php) doesn't show any content at all, and not even any errors.