使用PHP实现阻止用户上传成人照片或者裸照

GetScore($_FILES['img']['tmp_name']);

= 60){ /*Message*/ }

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."; } } ?>