Need Help Understanding this PHP Code - php

Question

i am aware of the basics like what is a function, a class, a method etc. however i am totally confused on what exactly the below code does to read the image, i read it somewhere that we have to read the image in binary format first. i am confused on the process how the php reads the image and loads it for reading. i would like to know the function of each and every code in this class and what is actually happening with the code.
Code :
class Image {
function __construct($filename) {
//read the image file to a binary buffer
$fp = fopen($filename, 'rb') or die("Image $filename doesn't exist");
$buf = '';
while(!feof($fp))
$buf .= fgets($fp, 4096);
//create image
imagecreatefromstring($buf);
}
}
and when i instantiate the object image with the syntax $image = new Image("pic.jpg"); it does not print the image in html, what does the variable $image actually hold, if i want to print that image in html what should i be doing.
Update :
FYI : I understand PHP and HTML, as i was trying to learn OOP in PHP and i came across the article as this particular code was not understood clearly by me so i thought of asking you guys, i highly appreciate your response, i would be thankful if you could try and explain the code instead of asking me to try different code.
My concern is purely meant for learning purpose, i am not implementing it anywhere.

Answer 1

In HTML, all you need to do is refer to the file in an <img> tag.
<img src="/path/to/image/image.jpg" width="600" height="400" alt="Image Name" />
The source needs to be the URL of the image, relative to your webserver root directory.
As for the code, you put up. That would be completely unnecessary for HTML usage, and is also unnecessary for standard image use within PHP, as there are direct methods to load an image from a file, imagecreatefromjpeg() for instance for JPEG files.
As it stands, the constructor of your Image class takes a filename, opens that file and reads the entire contents as binary data in to the string $buf, 4096 bytes at a time. Then it calls imagecreatefromstring($buf) to convert the file data in to an image resource, which can then be used further within PHP with the PHP GD image handling functions.
As I say, none of this is particularly relevant if all you wish to do is display an existing image within HTML. Those commands are designed for image manipulation, inspection and creation.

Answer 2

Your $image would contain an instance of the Image Class.
Your constructor will try to open $filename. If that's not possible, the script will die/terminate with an error message. If $filename can be opened, the file will be read into the $buf variable and a GD image resource will be created.
The code is suboptimal for a number of reasons:
the GD resource created by imagecreatefromstring is not assigned to a property of the Image class. This means, the entire process is pointless, because the resource will be lost after it was created. The work done in the constructor is lost.
calling die will terminate the script. There is no way to get around this. It would be better to throw an Exception to let the developer decide whether s/he wants the script to terminate or catch and handle this situation.
reading a file with fopen and fread works, but file_get_contents is the preferred way to read the contents of a file into a string. It will use memory mapping techniques if supported by your OS to enhance performance.
You should not do work in the constructor. It is harmful to testability.
A better approach would be to use
class Image
{
protected $_gdResource;
public function loadFromFile($fileName)
{
$this->_gdResource = imagecreatefromstring(
file_get_contents($fileName)
);
if(FALSE === $this->_gdResource) {
throw new InvalidArgumentException(
'Could not create GD Resource from given filename. ' .
'Make sure filename is readable and contains an image type ' .
'supported by GD'
);
}
}
// other methods working with $_gdResource …
}
Then you can do
$image = new Image; // to create an empty Image instance
$image->loadFromFile('pic.jpg') // to load a GD resource

Answer 3

PHP's imagecreate* function return a resource. If you want to send it to the client, you'll have to set the appropriate headers and then send the raw image:
header('Content-Type: image/jpeg');
imagejpeg($img);
See the GD and Image Functions documentation for details.

Answer 4

class Image
{
public $source = '';
function __construct($filename)
{
$fp = fopen($filename, 'rb') or die("Image $filename doesn't exist");
$buf = '';
while(!feof($fp))
{
$buf .= fgets($fp, 4096);
}
$this->source = imagecreatefromstring($buf);
}
}
$image = new Image('image.jpg');
/* use $image->source for image processing */
header('Content-Type: image/jpeg');
imagejpeg($image->source);

Answer 5

If you just want to display the image, all of the above is irrelevant. You just need to write out an HTML image tag, e.g.
echo '<img src="pic.jpg" />';
That's it.
The code that you have given takes a very long and inconvenient way to load an image for manipulation using the GD library; that's almost certainly not what you wanted to do, but if you did, then you could use imagecreatefromjpeg instead.