PHP 如何阻止用户上传成人照片或者裸照

程序员文章站 2022-05-23 20:43:02

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在这份教程中，我们将会学习到如何组织用户通过PHP上传成人照片或者裸照.

示例: http://www.rrpowered.com/demo/NudityFilter/
下载: http://www.rrpowered.com/code/RRPowered-NudityFilter.zip

我在phpclasses.org上面偶然发现一个很有用的，由Bakr Alsharif开发的可以帮助开发者基于皮肤像素点来检测图片裸照的类文件.

它会分析在一张图片的不同部分使用的颜色，并决定其是否匹配人类皮肤颜色的色调.

作为分析的结果，他会返回一个反映图片包含裸露的可能性的分值.

此外，他还可以输出被分析的图片，上面对使用给定颜色的肤色的像素进行了标记.

当前它可以对PNG，GIF和JPEG图片进行分析.

PHP

下面展示了如何使用这个PHP类.
让我们先从包含裸体过滤器，nf.php文件开始.

include ('nf.php');

接下来，创建一个新的名叫ImageFilter的类，然后把它放到一个叫做$filter的变量中.

$filter = new ImageFilter;

获取图片的分值并将其放到一个$score变量中.

$score = $filter -> GetScore($_FILES['img']['tmp_name']);

如果图片分值大于或等于60%，那就展示一条（告警）消息.

if($score >= 60){/*Message*/}

下面是所有的PHP代码:

/*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.";}}?>

标记语言

我们可以使用一个基础的HTML表单上传图片.

form method="post" enctype="multipart/form-data" action=" ">Upload image: input type="file" name="img" id="img" />input type="submit" value="Sumit Image" />form>

via:
http://www.oschina.net/translate/prevent-uploads-of-adult-or-nude-pictures-using-php
http://www.rrpowered.com/2014/04/prevent-uploads-of-adult-or-nude-pictures-using-php/