im trying to display the picture saved in the database using Php, but what i get is Url link which was saved in the database and not the picture, i think the syntax is wrong.
You need to construct an <img>
element, not just output the image URL.
Try replacing this: $row['ImageURL']
With this: "<img src='".$row['ImageURL']."' />
On this line:
echo "<td>" .$row['ImageURL'] ."</td>";
it looks like you'll want that to be:
echo "<td><img src='" .$row['ImageURL'] ."' /></td>";
Change
echo "<td>" .$row['ImageURL'] ."</td>";
to
echo "<td><img src='" .$row['ImageURL'] ."' style='width: 200px; height: 350px;' /></td>";
Modify the values 200px and 350px to whatever you want, just keep the 'px' suffix.