As I can see there are too many answers which are not accurate enough, so here goes mine:
This will print the image as you are doing it now(by the time of asking this question). As alternative to answer by @Vasil Dakov you should modify the snippet i gave you like this:
<?php
// ... Image generation goes here
header("Content-Type: image/jpeg");
ob_start();
print $im->getImageBlob();
$the_outputted_image = ob_get_flush();
?>
// Assuming that you use MVC approach and you are storing $the_outputted_image in a object and passing it to the view(ie. index.html or the HTML below the code).
//... Html code of index.html
<img src="data:image/jpg;base64 <?php print $the_outputted_image; ?>" alt="image" title="IMagick Generated Image" />
As another alternative is creating a script to generate the image, save it in some folder ( assuming img/ is the folder) and return only the path+filename+ extension to the file:
<?php
// ... Image generation goes here
header("Content-Type: image/jpeg");
$filename = 'img/' . md5(microtime()) . '.jpg'// Microtime is just as an example, you should use your own method.
$fp = fopen($filename, "x"); //Creating and opening the file for write-only
$im->writeImageFile($fp); //Writing the image to the file pointer (I would recommend writing it using, fwrite(), because it is binary-safe writing method)
fclose($fp);
?>
// Html
<img src="<?php print $filename; ?>" alt="image" title="IMagick Generated Image" />
documentation for Imagick::writeImageFile