How to return image - php

Question

I want to show image with php script. Here is my current code:
<?php
if (isset ($_GET['id'])) $id = $_GET['id'];
$t=getimagesize ($id) or die('Unknown type of image');
switch ($t[2])
{
case 1:
$type='GIF';
$img=imagecreatefromgif($path);
break;
case 2:
$type='JPEG';
$img=imagecreatefromjpeg($path);
break;
case 3:
$type='PNG';
$img=imagecreatefrompng($path);
break;
}
header("Content-type: image/".$type);
echo $img;
?>
But it doesn't show the image. What is the right way instead of echo $img?

Answer 1

just like this :
public function getImage()
{
$imagePath ="img/wall_1.jpg";
$image = file_get_contents(imagePath );
header('content-type: image/gif');
echo $image;
}

Answer 2

header("Content-type: image/jpeg");
imagejpeg($img);

Answer 3

There are functions for each format.
PNG: imagepng http://php.net/imagepng
JPEG: imagejpeg http://php.net/imagejpeg
GIF: imagegif http://php.net/imagegif
In your case you could add:
switch ($t[2])
{
case 1:
imagegif($img);
break;
case 2:
imagejpeg($img);
break;
case 3:
imagepng($img);
break;
}

Answer 4

I've used echo() for this before and it's worked. But try using imagejpeg() function instead.
Also, make sure you don't have any other content being output either before or after the image in your script. A common problem is blank spaces and line feeds being output caused by space and line feeds before or after the <?php and ?> tags. You need to remove all of these. And check any other PHP code loaded via include() or requre() for the same thing.