$dbc=mysqli_connect('localhost','root','','db')
$query="SELECT Binfile, Filetype FROM imagetable where imgid='$imgid'" ;
$result=mysqli_query($dbc,$query);
$row=mysqli_fetch_assoc($result);
$image=$row['Binfile'];
header("Content-type:".$row['Filetype']);
echo $image;
This is the code which I am using to display the image. But it is giving an error like
The image “http://lacalhost//viewimage.php?imgid=1”
cannot be displayed because it contain errors.
Help me to remove this error and display the image…
I wonder if I can use as echo'img src='
type to display.