Block Uploads of Adult or Nude Images using PHP

Block Uploads of Adult or Nude Images using PHP
I have created the File Uploading Script using a PHP class, that helps to detect image nudity based on skin pixel score developed by me. This code helps you to block adult or nudity images.Sample database design for Users.


Users

Contains user details username, password and email etc.

CREATE TABLE `users` (
`uid` int(11) AUTO_INCREMENT PRIMARY KEY,
`username` varchar(255) UNIQUE KEY,
`password` varchar(100),
`email` varchar(255) UNIQUE KEY,
`profile_image` varchar(255)
)

Sample values.

INSERT INTO `users` 
(`uid`, `username`, `password`, `email`) 
VALUES 
('1', 'akhan', MD5('password'), '[email protected]');

Javascript Code


$(“#photoimg”).on(‘change’,function(){})photoimg is the ID name of INPUT FILE tag and $(‘#imageform’).ajaxForm() – imageform is the ID name of FORM. While changing INPUT it calls FORM submit without refreshing page using ajaxForm() method. Uploaded images will <i>prepend</i> inside <i>#preview</i> tag.

<script type="text/javascript" src="http://ajax.googleapis.com/
ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript" src="jquery.wallform.js"></script>
<script type="text/javascript">
$(document).ready(function() 
{ 

$('#photoimg').on('change', function() 
 {
var A=$("#imageloadstatus");
var B=$("#imageloadbutton");

$("#imageform").ajaxForm({target: '#preview', 
beforeSubmit:function(){
A.show();
B.hide();
}, 
success:function(){
A.hide();
B.show();
}, 
error:function(){
A.hide();
B.show();
} }).submit();
});

}); 
</script>

 

Here hiding and showing #imageloadstatus and #imageloadbutton based on form upload submit status.

index.php
Contains simple PHP and HTML code. Here $session_id=1 means user id session value.

<?php
include('db.php');
session_start();
$session_id='1'; // User login session value
?>
<div id='preview'>
</div>
<form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'>
Upload image: 
<div id='imageloadstatus' style='display:none'><img src="loader.gif" alt="Uploading...."/></div>
<div id='imageloadbutton'>
<input type="file" name="photoimg" id="photoimg" />

</div>
</form>

 

ajaximage.php
Contains PHP code. This script helps you to upload images into uploads folder. Image file name rename into timestamp+session_id.extention

<?php
include('db.php');
session_start();
$session_id='1'; // User session id
$path = "uploads/";

function getExtension($str)
{
$i = strrpos($str,".");
if (!$i)
{
return "";
}
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}

$valid_formats = array("jpg", "png", "gif", "bmp","jpeg","PNG","JPG","JPEG","GIF","BMP");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
$name = $_FILES['photoimg']['name'];
$size = $_FILES['photoimg']['size'];
if(strlen($name))
{
$ext = getExtension($name);
if(in_array($ext,$valid_formats))
{
if($size<(1024*1024)) // Image size max 1 MB
{
//---Image Filter Code
require_once('class.ImageFilter.php');
$filter = new ImageFilter;
$score = $filter->GetScore($_FILES['photoimg']['tmp_name']);
if(isset($score))
{
if($score >= 60) // Score value If more than 60%, it consider as adult image. 
{
echo "Image scored ".$score."%, It seems that you have uploaded a nude picture :-(";
}
else
{
//---Image Filter Code 
$actual_image_name = time().$session_id.".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name))
{
mysqli_query($connection,"UPDATE users SET profile_image='$actual_image_name' WHERE uid='$session_id'");
echo "<img src='uploads/".$actual_image_name."' class='preview'>";
}
else
echo "failed";
//---Image Filter Code 
}
}
//---Image Filter Code 
}
else
echo "Image file size max 1 MB"; 
}
else
echo "Invalid file format.."; 
}
else
echo "Please select image..!";
exit;
}
?>

 

db.php

Database configuration file, just modify database credentials.

<?php
error_reporting(0);
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'username');
define('DB_PASSWORD', 'password');
define('DB_DATABASE', 'database');
$connection = @mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
?>

 

If you have any questions do let me know in comments section below 🙂

Share your thoughts
Ahtasham Khan