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.
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.
|
1
|
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.
|
1
|
$filter = new ImageFilter; |
获取图片的分值并将其放到一个$score变量中.
|
1
|
$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.
|
1
2
3
|
if($score >= 60){/*Message*/} |
下面是所有的PHP代码:
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
|
<?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.
|
1
2
3
4
5
|
<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> |
