I create php script that shows the Id and the image stored in mysql database, but i can just see the image without the Id, how can i show the image and the id in this php script ?
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image,image_type FROM images where id = $id";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
$ss=$result['image_type'];
if ($ss == 'php') {
echo 'Type of image ' . $result['image_type'];
echo ( $result['image']);
} else if ($ss == 'android') {
echo 'Type of image ' . $result['image_type'];
echo base64_decode( $result['image'] );
}
mysqli_close($con);
}
?>
this is what i get when i execute script :
note : when i remove echo 'Type of image ' . $result['image_type']; from the code the image is properly displayed. but i want to show the image and its type too.