PHP 如何阻止用户上传成人照片或者裸照
PHP 如何阻止用户上传成人照片或者裸照
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:
<?php /*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="<?php echo $SERVER['PHP_SELF'];?> "> 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/