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I would like to have code for an image that counts the number of times the image is viewed, regardless of what site the image is found on. I want to use the img src tag, and have src point to a php page that counts that view and then returns the image to be viewed. I was thinking something like this: 

 <img src="www.mywebsiteurl.com/something.php" />

How would I go about writing "something.php" so that it returns the proper image file? I know how to record the page view, but if I must do something different in this case, please tell me.

Lightness Races in Orbit
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Programmer Person
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6 Answers6

2

Your script needs to:

  1. Record the hit.
  2. Set the content type of the output ( header('Content-Type: image/jpeg'); )
  3. Output the image ( readfile( $path_to_image ) );

For examples, please see the readfile documentation and Output an Image in PHP.

Community
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George Cummins
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1

Loading images / files from php is slow than normal loading files.

Main cons:

  • Images can't be cached due to dynamic query most browser dont support on dynamic queries.
  • Overhead on sever and apache both.

If you use database and want to increment column after image loads than a person can load your image again and again. So you need to use conditions on every query.

Alternative Solution

Enter this code in footer of page on which you want to show images

A ajax call

$.get("inc_img_view.php?img_id="+<?php echo $img_id; ?>);
Ravinder Payal
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0

This is the most barebones version you can have:

<?php

header('Content-type: image/jpeg');
readfile('somepic.jpg');
Marc B
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0

You can call an image using a URL similar to: www.mywebsiteurl.com/image_renderer.php?url=url_to_image.

image_rendered.php:

<?php
header("Content-Type: image/jpeg");
$url = urldecode($_GET['url']);

$headers = array(  
"GET ".$url." HTTP/1.1",   
"Content-Type: text/xml; charset=UTF-8",  
"Accept: text/xml",
"Cache-Control: no-cache",  
"Pragma: no-cache"
);

$ch = curl_init($url);  
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE);  
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);  
curl_setopt($ch, CURLOPT_POST, false);  
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers); 
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, false);

$r = curl_exec($ch);

// you can connect to your database and increment the number of times this image was viewed

echo $r;
?>

Hope this will point you into the right direction.

Artem Kalinchuk
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Yes you're on a good start. I'd do something like this

<img src="www.mywebsiteurl.com/image_counter.php?src=image.jpg" />

In image_counter.php, take $src = $_GET['src'] for image source.

Then +1 the hit for that image in the database.

Then do

header('Content-type: image/jpeg');
readfile($src);
Moe Sweet
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0

Something like this should work:

<?php

$file = $_GET['img'];
$path = "images/";

// do an increment of views on the image here - probably in a database?

$exifImageType = exif_imagetype($path.$file);
if ($exifImageType !== false) {
    $type = image_type_to_mime_type($exifImageType);
}

header('Content-Type: $type');
echo file_get_contents($path.$file);

Note: this would make something like:

<img src="getimage.php?img=image.jpg" />

show the correct image after incrementing the views.

You could always add some .htaccess to make all calls to /images/anything go through your image file. It'd mean you could keep the URL looking as though it's getting an actual image file (e.g. /images/logo.jpg would be re-written to actually go through getimage.php?img=logo.jpg)

Nick
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