PHP identifies pornographic images uploaded by users and prevents instances-PHP source code

In this tutorial, we will learn how to use PHP to identify adult photos or bare images uploaded by users. The principle is to analyze whether the color of an image is similar to that of the human skin in a large area. In this tutorial, we will learn how to use PHP to identify adult photos or bare images uploaded by users. The principle is to analyze whether the color of an image is similar to that of the human skin in a large area.

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On phpclasses.org, I accidentally found a very useful file. Developed by Bakr Alsharif, it can help developers detect bare image files based on skin pixels.

It analyzes the colors used in different parts of an image and determines whether it matches the human skin color.
As a result of the analysis, the system returns a score that reflects the possibility of image exposure.
In addition, he can output the analyzed image and mark the pixels with the color given.

Currently, it can analyze PNG, GIF, and JPEG images.

PHP

The following shows how to use this PHP class.

Let's start with the inclusion of the native filter, nf. php file.

The Code is as follows:
Include ('nf. php ');

Next, create a new class named ImageFilter and put it in a variable named $ filter.

The Code is as follows:
$ Filter = new ImageFilter;

Get the image score and put it in a $ score variable.

$ Score = $ filter-> GetScore ($ _ FILES ['img '] ['tmp _ name']);

If the image score is greater than or equal to 60%, an alarm message is displayed.

The Code is as follows:
If ($ score> = 60 ){ /* Message */ }

Below are all the PHP code:

The Code is as follows:

/* Include the Nudity Filter file */ Include ('nf. php '); /* Create a new class called $ filter */ $ Filter = new ImageFilter; /* Get the score of the image */ $ Score = $ filter-> GetScore ($ _ FILES ['img '] ['tmp _ name']); /* If the $ score variable is set */ If (isset ($ score )){ /* If the image contains nudity, display image score and message. Score value if more than 60%, it is considered an adult image .*/ If ($ score> = 60 ){ Echo "Image scored". $ score. "%, It seems that you have uploaded a nude picture ."; /* If the image doesn' t contain nudity */ } Else if ($ score <0 ){ Echo "Congratulations, you have uploaded an non-nude image ."; } } ?>

Markup Language

We can use a basic HTML form to upload images.

The Code is as follows:

Summary

Please remember that PHP cannot detect all naked pictures, so it is not completely credible. I hope you will find this useful.